Quadratics

May 20th, 2017

J: What do you know about quadratics?

S: I know that there is the quadratic formula.  I recently did a lot of quadratics, but I don’t remember exactly what I did.  Let’s just start with a problem using the quadratic formula.  Oh!  I just realized that I memorized it!

(-B ± √-B² – 4AC) ÷ 2A

Now, give me an equation to solve.  And remember: This is Beatles math.

J: George is designing an amplifier horn.  It will be in the shape of a parabola, and the 0,0 origin on the blueprints will be offset for mounting.  George will need to know the two places where the right and left of the amplifier horn are at elevation y = 0.  The equation for the horn is y = x² – 4x + 1.

S: So, I’m trying to get it where y = 0?

J: Yes.

S: Uhhhhhhhhhhhhhh…

J: So, a quadratic equation is in the form ax² + bx + c.

S: Oh!  I remember that!  So, how does that help me solve it?

J: Well, Paul quips that.

S: Quips what?

J: So, the thing I said was in quotes, and after that, the text reads, “Paul quipped”.

S: Oh.  So: ‘”So, a quadratic equation is in the form ax² + bx + c,” Paul quipped.’  Still, how does that help me?

J: Oh, well, Ringo says, “So, for Geo’s equation, ‘a’ would be what?”

S: Oh!  It would be 1.

J: John said.

S: But I said that!

J: “And ‘b’ would be…?” *bad impression of Ringo’s voice*

S: -4.

J: “And ‘c’?”

S: 1.  Having the Beatles teach me math in this story makes it harder to concentrate.  But I want them to keep doing it!  But I feel so weird and fangirlish!

J: And John says, “So, we plug a, b, and c into the quadratic formula.”

S: Oh!  Now I remember how to do everything!  Thanks for jogging my memory, Beatles!

a = 1

b = -4

c = 1

(4 ± √4² – 4 · 1 · 1) ÷ 2 · 1

I’ll solve it with the + first.  *solves it*  That equals 3.7320508075688…

Now for the -.  *solves it*  That equals 0.267949192431…

J: Shall we check those?

S: Okay.

J: So, we could punch into the calculator “(4 + √12) ÷ 2”.

S: And the answer is correct!

J: And now the other one.

S: Also correct.

J: Those two numbers are the roots of the equation, so we can write the equation as y = (x – 3.73) · (x – 0.26).

S: And we still have to make that 0?

J: “We have to make y = 0,” Paul says.  “So, y is 0 when the first subexpression is 0 or when the second subexpression is 0.”

S: Oh!  Because anything multiplied by 0 is 0!

J: “Roight.”

S: This doesn’t make sense because we have no value for x!

J: Ringsy says, “What minus 3.73 is equal to 0?”

S: 3.73.

J: Correct!  And, for the other x, what minus 0.26 is equal to 0?

S: 0.26.

J: Correct!  George finally says, “Thank you, boys!  So, y = 0 when x = 3.73 and 0.26.”

S: Happy endings = gear

J: So, can we graph it?

S: First, what exactly is an amplifier horn?

J: Just an amplifier shaped like a horn.

S: But why is George building one when he could just buy one?

J: Maybe he’s doing it to make an Indian swami happy.  *holding phone up*  Here’s an amplifier horn for an iPhone.

S: Disturbing.  Anyway…graphing.

J: You could do some algebra to figure out when x = 0.  But that’s optional.  Do you want to do it?

S: No, I think we’re out of time.  But why did George need to know the…things for the equation for the parabola-shaped amplifier horn when he could have just drawn a parabola?

J: Because his friend, who’s going to machine it, prefers a parabolic equation.

S: Ah.  Toodles!

Functions

MAY 14TH, 2017

S: First, we should do the table functions, and then the graph functions.  Just an overview for those of you who don’t know: The goal of solving a table function is to find an equation that, when you plug in x, always equals y.  The table gives examples of what y is when x is different values.  A graph function is an equation graphed.

J: John writes a table for Ringo to solve.

x	y
0	0
1	1
2	4
13	169
1,024	1,048,576

What equation does Ringo come up with?

S: x²

J: John makes a harder table for Ringo to solve, as Ringo seems smarter than John thought.

x	y
0	1
1	2
2	7
3	16
4	29
5	46

Ringo gives the problem set to George.

S: George is eating an egg sandwich and doesn’t feel like doing algebra, so he gives it to Paul.  Paul solves it, but he doesn’t know that John made it, so he asks John if John can solve it just to see if he can.  And John concludes from this that Ringo cannot solve the problem.  So, I will solve it.  I will re-write the table and think of the possible answers for each individual pair of numbers, and then see if any of them match.

x	y	GUESSING
0	1	+ 1
1	2	+ 1
2	7	+ 5 · 2 + 3
3	16	· 2 + 10
4	29	· 7 + 1
5	46	· 10 – 4

Gaaaaaah!

J: You might try a column of x² and see how it looks.

S: Okay.

*ends up trying many more columns that Jeff suggests, but only gets to Nowhere Land*

x	y	x2	DIFFERENCE BETWEEN Y & X2	DIFFERENCE BETWEEN X2 AND DIFFERENCE BETWEEN Y & X2
0	1	0	1	1
1	2	1	1	0
2	7	4	3	1
3	16	9	7	2
4	29	16	13	3
5	46	25	21	4

J: Do you want a hint?

S: Yes.

J: Do you see the last column kind of matching some other column?

S: Sort of.  It matches the x column.

J: So, those differences – it’s looking like x – 1 = the difference between x² and the difference between y & x².

S: Except for the first row.

J: Sure.

S: How does that help?  The last column isn’t even y!

J: Sure.  Try a column now where you add that back into the output (y).

S: That makes no sense.  Can I just try by myself?

x² – 1 = k

0² = 0 – 1 = -1, x² + 1

1² = 1 – 1 = 0, x² + 1

2² = 4 – 1 = 3, x² + 3

0² = 9 – 1 = 8, x² + 7

0² = 16 – 1 = 15, x² + 13

0² = 25 – 1 = 24, x² + 21

I cannot do this.  Help!  I need somebody (A.K.A. you)!

J/S:

x	y	x2	DIFFERENCE BETWEEN Y & X2	DIFFERENCE BETWEEN X2 AND DIFFERENCE BETWEEN Y & X2	Y + DIFFERENCE BETWEEN X2 AND DIFFERENCE BETWEEN Y & X2
0	1	0	1	1	2
1	2	1	1	0	2
2	7	4	3	1	8
3	16	9	7	2	18
4	29	16	13	3	32
5	46	25	21	4	50

S: If you compare x to the last column, the equation 2x² works, but only if we pretend that the last column is the output (y), which it isn’t, and only if there wasn’t that stupid zero row that doesn’t match!!!

J: Forget about the zero case.  Zero cases are often the exception.

S: I don’t want to do this anymore!  You should have told me in the beginning that x = 0 didn’t matter!!!

J: Well, shall we go on fast-forward, then?

S: If you put the Beatles in on the next line.

J: Paul points out that y appears to be equal to (y + difference between x² and difference between y & x²) – (difference between x² and difference between y & x²).

S: OHHHHHHHHHHHHHH!!!

50 – 4 = 46

32 – 3 = 29

18 – 2 = 16

8 – 1 = 7

2 – 0 = 2

J: The last column is potentially equal to 2x².  The second-to-last column looks like x – 1.  And so, y appears to be equal to 2x² – (x – 1), or 2x² – x + 1.

S: That’s the equation.  Don’t ask me how it was done.  I can’t do that ever again.  Time to  check to see if it matches, even the zero case.

2x² – x + 1 = y

2 · 0² – 0 + 1 = 1

2 · 1² – 1 + 1 = 2

2 · 2² – 2 + 1 = 7

2 · 3² – 3 + 1 = 16

2 · 4² – 4 + 1 = 29

2 · 5² – 5 + 1 = 46

Everything matches!

Time for part two of the lesson: graphing the function…

*Realizes that graph is terrible because it is not A, even, B, big enough, or C, correct*

J: Where’s your graph paper?

S: I’ll get it…

That is a slightly better graph.  Grand finale time!  Well, what is the grand finale?

J: Paul kisses you, saying, “Thank you!  This is the perfect parabolic hat for my next show!”

S: “Parabolicat”???

J: “Parabolic hat“!

S: Oh.

J: “ParaboliCAT”? HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA

S: As you laugh, your daughter is dead.

J: NOOOO!!!  Did the parabolicat eat her?

S: No!  I died of joy because Paul kissed me!

J: Oh.

S: Can’t wait to see you in heaven!  It’s nice up here.

J: So, now you’re up there with John and George, waiting for Paul and Ringo.

S: No!  Because John created the problem for the parabola, and I solved it on the same day!  I went back in time to 1964 or something and died, so they’re all still alive!

J: And you disrupted my conception, so I was never born!

S: And I was never born, so I never died!  Oh, that’s sad.

J: But John never died, and neither did George!  You saved the Beatles!!!

S: No!  It’s still 1964!  Except you’re in the present day, even though you don’t technically exist!  Not even Brian is dead yet!

J: Yes, but you disrupted the timeline enough that they’re all still alive in the present day!

S: Really?

S:

S:

S:

S: YAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAY!!!

I WILL ALWAYS BE REMEMBERED AS THE DEAD FANGIRL WHO SAVED THEM!

But I was never born, and they never died…

Whatever!

At least I know that I saved them, and at least Paul kissed me!🤣

Two Inequalities, Two Unknowns

April 29th, 2017

J: The additive inverse of a number is the negative of that number.  We also have the multiplicative inverse.  The multiplicative inverse of a number is its reciprocal.

S: Okay.

J: So, if we have an equality such as 3x = 4y, we could take the inverses of both sides, and the equality is still true.

S: Let me try that.  But which inverse?

J: Either one or both!

S: 1/(3x) = 1/(4y)

J: That would be the inverse.

S: Dad, just a warning: If the Beatles aren’t in here in two lines, I’m quitting.

J: I’ll see what I can do.  So, that would actually be 1/(3x) = 1/(4y).

S: I wrote what my dad said in his previous line, when he said: “Next line”, but then I had to almost quit because I did go to the next line.  I told him he could get a second chance if he erased “next line” and said something Beatle-y.

J: John erases this.  *erases “next line”*

J: George tickles Starli.

S: AAAAAAAAAA-HA-HA-HA *is being tickled*

J: Okay, so, starting over.  Paul is considering playing Starli’s Yamaha keyboard, but wants to think it over.

S: Because it would immediately become A, very valuable, and B, an heirloom.  And why is he here? *slight fangasm imagining him actually being here and playing a song on my keyboard*

J: Because Starli insisted on Beatles in her math, so it takes ten times as long to learn the same thing.

S: HA.     HA.     HA.     HA.     VERY FUNNY.     NOT.

J: Paul’s piano has 88 keys.  And Starli’s piano has…?

S: 77.

J: Write an inequality involving those two numbers.

S: Okay.  88 > 77.

J: Now take the inverse of both sides of that inequality.  Your choice: multiplicative or additive.  Is it still true?

S: 1/88 > 1/77.  It is not true.

J: How can you fix it?

S: Flip the sign!  1/88 < 1/77.

J: Hmm… So, you flipped the inequality symbol, right?

S: Yes.

J: Good.  Now, write another inequality in terms of those numbers of keys.

S: 77 < 88.  There.

J: Now, multiply both sides by -1, which is the same as taking the additive inverse.

S: 77 · -1 < 88 · -1 ≠ -77 < -88!!!  Flip the sign.  -77 > -88.

J: So, we can conclude from this that when you’re solving an inequality and you’re taking the inverse of both sides, you need to flip the inequality sign.

S: Which I already knew!

J: So, what happens if you have 88 > 77 and you took both the multiplicative inverse and the additive inverse?

S: 88 · -1 > 77 · -1 ≠ -88 > -77 ≠ 88 > 77, but that is equal to its original self.  So, people, if someone asks you to take an even number of inverses, don’t change the equation at all!  Lesson learned; done.

J: Oh.  No.  Don’t change the inequality symbol.

S: But, what I said is also true because… it’s true.

J: No!  If we’re trying to solve an equation by the methods we learned in other lessons, we’ve probably done a multiplicative or additive inverse before.  And when we’re doing this when solving an inequality, we have to be mindful of flipping the greater than/less than symbol.

S: So, we’re both right!!!  Because if you do both, it doesn’t change the number or the inequality sign!

J: If we did both, the numbers are different!

S: Oh!  I see!  Back when I was going to do both inverses, I only did one inverse!!!  So, it would actually be -1/88 > -1/77.

J: Is it true?

S: Well, it’s weird, because there’s a fraction and it’s negative, so the regular wrongness cancels out.  But yes, it’s true.

J: Are we done with that part?

S: Yes.  So, in one sentence, what was the takeaway from that part of the lesson?

J: If you take an inverse, make sure to flip the inequality symbol, but if you take an inverse an even number of times, don’t flip it, but do flip it if you take an inverse an odd number of times.

S: Okay!  Two inequalities, two unknowns, FOUR BEATLES WHO HAVE NOT COME UP IN 28 LINES…  Actually, Ringo hasn’t been mentioned at all!

J: It’s not his time, though, unless you want it to be his time.

S: It’s always his time!  And we need the story problems to be more complicated and interesting so that our Beatlemaniacal audience will be entertained!

J: Okay, well, George and Ringo have been eating mini sandwiches from platters passing by at a party.

S: Oh, that reminds me of that scene in “A Hard Day’s Night”…

J: They’re not sure how many they’ve eaten, but they know a little bit about how many they’ve eaten relative to each other, and they know approximately how many they’ve eaten, because a mathematician friend has been observing and has provided us with two inequalities to solve together.

S: And “we” are Mona and Pete Best.  May I ask where John and Paul are?

J: Having tea with the Queen.

S: Wow.  George and Ringo must feel unhappy and left out.

J: They missed the flight.

S: So George and Ringo aren’t in England?

J: They’re at the fair in Edinburgh.

S: Anyway, John and Paul and Brian and whoever else WOULD CHECK TO MAKE SURE THAT THEY WERE ALL ON THE PLANE.

J: They are all working for MI6, so they’re going undercover and role-playing and sacrificing their usual Beatle-brotherhood for the good of Queen and country.  So, shall we use abbreviations?

S: No.

J: 3george – 2ringo ≥ 5 sandwiches

ringo – 2george ≤ -4 sandwiches

So, use the system of inequalities to find out how many sammiches they ate.

S: “Sammiches”?

J: Yes, sammiches, made by the Earl of Sandwich for their express culinary pleasure.

S: Isn’t there a town called Sandwich in England?

J: *Googles it*  Yes.  It has a population of 4,985.

S: George should live there, shouldn’t he?

J: Of course.

S: So, solving the inequalities…

3george/3 ≥ (5/3 sandwiches + 2ringo)/3

george ≥ 5/3 sandwiches + 2/3ringo

ringo – 2 (5/3 sandwiches + 2/3ringo) ≤ -4 sandwiches

ringo – 2 · 5/3 sandwiches + 2 · 2/3ringo ≤ -4 sandwiches

ringo – 3.33 sandwiches + 1.33ringo ≤ -4 sandwiches

2.33ringo – 3.33 sandwiches ≤ -4 sandwiches

2.33ringo ≤ -4 sandwiches + 3.33 sandwiches

2.33ringo/2.33 ≤ -0.66/2.33 sandwiches

ringo ≤ -0.28 sandwiches

REALLY, DAD?  Ringo ate -0.28 sandwiches?!  Oh!  Ringo ate less than or as much as -0.28 sandwiches.

J: That’s not the result I was looking for.

S: oh no OH NO OH NO

J: Shall I take another look at the inequalities?

S: Okay.

J: *reviews inequalities for ten minutes*  This doesn’t make sense…  Your inequalities are correct, but I know that Ringo ate two sandwiches and George ate three, because that’s what I based the inequalities off of…  *reviews inequalities for another ten minutes*  Oh, I see!  You dropped a sign when you were distributing the -2.  5/3 sandwiches was multiplied by -2, but 2/3ringo was multiplied by 2.

S: So I have to redo it.  Wonderful.  But WAIT!  I know the answer!

J: No, no, no!  It would be un-Beatlesque of you to cheat!

S: You were the one who carelessly told me the answer!

J: But you don’t get to use the answer!

S: Alright, I’ll start with the wrong inequality and write it correctly.

ringo – 2 · 5/3 sandwiches + -2 · 2/3ringo ≤ -4 sandwiches

ringo + -3.33 sandwiches + -1.33ringo ≤ -4 sandwiches

-0.33ringo + -3.33 sandwiches ≤ -4 sandwiches

-0.33ringo ≤ -4 sandwiches + 3.33 sandwiches

-0.33/-0.33ringo ≤ -0.66/-0.33 sandwiches

ringo ≥ 2 sandwiches

Ringo ate two or more sandwiches.

3george – 2ringo ≥ 5 sandwiches

3george – 4 ≥ 5 sandwiches

3/3george ≥ ≥ 1/3 sandwich

George ate 1/3 of a sandwich.

Which is wrong.

J: Shall I have a look?

J/S: *both reviewing*

S: What the heck is going on?  I wrote two inequality signs and dumped the -4!  Okay…

3george ≥ 5 + 4

3/3george ≥ 9/3

George ≥ 3

George ate three or more sandwiches.

Our summary is that Ringo ate two or more sandwiches, and George ate three or more sandwiches.  Good job, Pete!

J: Anytime, Mona!

Two Equations, Two Unknowns

April 16th, 2017

S: Before we begin, I have two things to say to whoever is reading this.  1: I’m really sick and I’m losing my voice, so can’t talk, so I’m literally just talking on paper and my dad has to read what I write to know what I’m saying.  2: My dad was being really crabby about the Beatle number system, so, unfortunately, I have decided to stop using it.

J: Problem #1:

S: Problem #John!!!  Just kidding!  I promise I won’t do that again.

J: Problem #1: x is the number of songs that John wrote this month, and y is the number of songs that Paul wrote this month.

S: Which month?  If you’re talking about April 2017, keep in mind that it’s impossible for x to be anything other than zero.

J: The problem does not specify which month.

S: You’re making the problem up, though!

J: Are we ready?

S: 😡

S: Fine.

J: I can go to the other room and Google up some actual data.

S: Or you could make up a month.

J: Beatleary.

S: U NO WOT I MEANT!!!  Fine.  Just tell me the problem.

J: Well, the fan club newsletter-

S: -run by Freda Kelly-

J: -gives a report of two equations, two unknowns and asks the question, “Who is ahead for this month and by how much?” and…

x + 2y = 13

3x – y = 4

S: Those are really simple compared to what’s on School Yourself (School Yourself is where I do math, and it’s a wonderful learning resource!  Link: www.schoolyourself.org).  But I’ll do the problem.

x = 13 – 2y

3 · (13 – 2y) – y = 4

3 · 13 – 3 · 2y – y = 4

39 – 6y – y = 4

-6y – y = 4 – 39

-7y = -35

y = -35/-7

y = 5

Time to plug it into the other equation…

x = 13-10

x = 3

x = 3 and y = 5.

You did that on purpose because Paul is my favorite, didn’t you?

J: Yes.

S: Knew it.  Next problem!

J: It’s a different month.  Same problem.

S: WHAT???  Be creative!  We have an audience!

J: Beatlember.

S:

J: Is that not creative?

S:

J: Oh.  Our AUDIENCE.

J: George and Ringo met some girls.  The number of girls George met is x, and the number of girls Ringo met is y.  How many did they meet at the party in question all together?

S: Now.

J:

J:

J: x² + 2y = 6

3x + 2y² = 8

S: Somewhere here, I’ll have to use the quadratic formula.  Which I forgot.  Luckily, it’s written down.  Oh, and by the way, for those of you who don’t know the quadratic formula, it’s (-B ± √B² – 4AC) ÷ 2A.

J: You don’t actually have to use it.

S: Oh.  But you said I did!

J: I recommend just doing regular substitution, as you did in the last problem.

S: Whatever “regular substitution” means.

J: Get x all alone.  And then substitute x‘s value into the second equation and solve for y.

S: Okay.  I’ll write out the equation again and try to solve it…

x² + 2y = 6

3x + 2y² = 8

2y = 6 – x²

y = 6/2 – x²/2

y = 3 – x²/2

3x + 2 (3 – x²/2)² = 8

WOW.

3x + 6 – 2 · (x²/2)² = 8

6 – 2 · (x²/2)² = 8 – 3x

How do I get x out of the parentheses?

J: So, for taking the term in the parentheses to a power, we could simply use FOIL-

S: I hate FOIL.  Is that the only way?

J: The other way is to square the numerator and square the denominator.

S: I’ll do that.

6 – 2 · x^{2^2}/2² = 8 – 3x

J: I don’t think you’re solving it correctly.  When you distributed the 2, you forgot to keep the ² with the 3.

S: Okay, I’ll go back…  Once again, I have:

3x + 2 (3 – x²/2)² = 8

3x + 18 + 2 (x²/2)² = 8

J: That is not the same equation.  FOIL is the reality.  Multiply the (3 – x²/2) by itself.  Then multiply all of those terms by the 2.

S: Okay.  3x + (3 – x²/2) · (3 – x²/2) · 2 = 8

J: Correct.  Now we can FOIL the two terms in parentheses.

S: Alright.  I hate FOIL, but I’ll try.

F     3 · 3

O     3 · – x²/2

I     – x²/2 · 3

L     – x²/2 · – x²/2

3 · 3 = 9, and everything else is the same.  So, 9 + 3 · – x²/2 + – x²/2 · 3 + – x²/2 · – x²/2

J: Yeah.  So that can be further simplified.

S: How?

J: Well, you can put the numbers first.

S: So, 9 + – 3 · x²/2 + – 3 · x²/2 + – x²/2 · – x²/2

And 9 + – 6 · x²/2 + – x²/2 · – x²/2 = 8

And 3x + 2 (9 + -6 · x²/2 + x²/2 · x²/2) = 8

We’re back in parentheses.

J: And we distribute the 2.

S: Which I tried to do in the first place.

J: But it didn’t get to the first term; it only went to the first and then it went sideways.  By the way, this is an unusually hard problem.  I don’t think you did problems like this on School Yourself.

S: The inequalities seemed this hard.

J: Inequalities are hard for a different reason.

S: So, distributing the 2…

3x + 18 + 2 (-6 · x²/2) + 2 (x²/2 · x²/2) = 8

We’re back in those dumb parentheses!  And the Beatles haven’t come up in paragraphs and paragraphs!

J: Well, let’s simplify-

S: BEATLES!!!

J: -the 2 (-6 · x²/2), which is equal to -6x².

S: 3x– GAAAAAAH I’M SO SICK OF THIS I WANT TO SLEEP THE BEATLES ARE PRACTICALLY ABSENT FROM THIS WHOLE THING!!!

J: John is very sick and is having trouble solving this problem, and he is very impatient with Paul’s attempts to help him.

S: Okay.  What can we do about it?  We aren’t anywhere near that time or place.

J: We can simplify the equation further and try to find out how many girls George meets and hope that they will somehow peer through a black hole and see the answer.

S: Alright.  3x + 18 + -6x² + 2 x⁴/4 = 8

Okay.  18 + -6x² + 2 x⁴/4 = 8 – 3x

-6x² +  2 x⁴/4 = -10 – 3x

-6x² + x⁴/2 = -10 – 3x

J: I’d leave them on the other side.

S: 😠

J: Okay, I think it’s time to just take some guesses on how many girls George met.  He can’t have met 0 or a negative number.

S: He could have met 0.

J: So, let’s plug in 0 and see if the equality is true.

S: NO.  Because you said it wasn’t, and I don’t want to waste time because I am sick and tired.

-6 · 0² + 0⁴/2 = -10 – 3 · 0

We can just see that 0 doesn’t work because anything multiplied by 0 is 0, and then everything would just be 0.

J: How about 1?

S: That changes nothing, though.  Well, I guess…

-6 · 1² + 1⁴/2 = -10 – 3 · 1

-6 + 0.5 = -13

-6 + 0.5 ≠ -13

1 is incorrect!  I don’t want to go on!

J: You must!  Try 2 next.

S: I caaaaaaaaaan’t!!!

J: Please please me and try 2.

S: Fine.

-6 · 2² + 2⁴/2 = -10 – 6

-24 + 8 = -16

FINALLY!!!  GLORY IZ MINE!!!

J: You know how many young women George met.

S: -16.  Wait a second!

J: The equality is true when you plugged in 2, so 2 satisfies the equation.

S: Okay.  So, back to the two original equations, with x as 2:

2² + 2y = 6

2y = 6 – 2²

2/2y = 2/2

y = 1

George met two girls and Ringo met one.  John and Paul peered into the black hole and found the answer!  Now John can go to bed.

Dividing Fractions

April 8th, 2017

J: So, there are first the preliminaries.  So, 1 ÷ 1…

S: Equals 1.

J: Yes.  1 ÷ 2…

S: Equals 0.5.

J: Or 1/2.  Yeah.  Now 1 ÷ 0.5…

S: 2?  That makes no sense.  I’m afraid I don’t understand.

J: Right!  So, let’s talk about the power of -1.

S: I know that -1 is powerful.  Very powerful.  When does this get to the you-know-whos?

J: Right now.  First, 1^-1…

S: Is -1…

J: No, it equals 1.

S: Why?

J: Our general equation is x^-1 ≡ 1/x (the three lines mean “defined to be”).

S: Oh.  So 1^-1 is just 1 ÷ 1.  So, how does that show the power of -1?!  And I thought you said that we would get to the you-know-whos a long time ago.

J: Yes.  And we are halfway in getting to that right now.

S: *facepalm*

J: So, question: What is (2^-1)^-1?

S: Um, 0.5^-1…  1 ÷ 0.5 = 2.  Why is half of half of one equal to two?

J: So, obviously, half of half of one is a quarter, right?

S: But I think that 1 ÷ 0.5 should not be 2.  I think that it should be undefined because it makes no sense!

J: Well, let’s look at another little equivalence.  y/x = y · x^-1

S: Is this true for all math?

J: Yes, aside from the little problem of dividing by zero.

S: This has nothing to do with dividing fractions or the Fab Four.

J: It has EVERYTHING to do with dividing fractions!  But let me just tell you the graphical trick: If you’re dividing Paul by Ringo over George, just flip George/Ringo upside-down, and multiply Paul by that.

S: Paul ÷ Ringo/George, flipped is Paul · George/Ringo.

J: Yes!  So, another one: 3 ÷ 1/2 is the same as 1/2 flipped upside-down.

S: The thing that starts with an ‘r’.  A fraction’s r…

J: I can Google it.  *Googles it*  Reciprocal!

S: Okay, so its reciprocal.  Is this true for all math?

J: Yes!  Aside from the discontinuity of dividing by zero, yes.

S: We should have code names for numbers.  Zero is Pete Best.  One is John.  Two-

J: AAA!  BLEH!  So, this business says that the reciprocal of 1 is 1.

S: Because it’s a whole number.  1 and -1 is John, by the way, and we add a minus sign in front of his name when we are doing -1.  Same with Paul = 2, Geo = 3, and Ringo = 4.

J: Okay.  But it’s not just because it’s a whole number.  1 is in the middle of everything; it’s unity-

S: You mean John.

J: John is unity!  I am he as you are me and we are all together!  John is the only case where a number equals its reciprocal!  So, the reciprocal of 1/2 is 2/1, or 2.

S: And that’s why 1 ÷ 0.5 is 2.  Because that equation said… reciprocal?  No, it didn’t.  Whatever!  I get it.  I need to actually divide fractions now.

J: Okay!  2/3 ÷ 5/7

S: That is not the Beatles.  They have not been in here enough.  Oh!  We cannot say 2 or 3, and up there in my last quote, it should have been John ÷ Pete.5 is Paul.

J: I like Beatles math – I really am scratching my head for a good example here – but I really am concerned that if you don’t see the actual numbers graphically, you won’t understand the concept.

S: I know that John is 1.  It is  1.  For example, how many dots do I see?  …  1, 2, 3; I mean John, Paul, George dots.

J: It’s more interesting if we use prime numbers like 2, 5, and 7.

S: I think you mean Paul, 5, and 7.

J: BLEEEAH!

S: How hard is it to remember five little numbers?

J: Fine!  Let’s do THIS!!!     😤     John/Paul ÷ George/Ringo

S: Well, I still don’t know how to divide fractions, which you were supposed to teach me!

J: Bleeeah!

S: So, John/Paul ÷ George/Ringo = John/Paul · the reciprocal of George/Ringo, which is Ringo/George.

J: Could we rewrite that?

S: John/Paul · Ringo/George.  Oh!  John · Ringo = Ringo, and Paul · George = 6.  Ringo!  That is the answer, also equivalent to Paul/George.

J: Can we just do a numbers one?

S: Those are numbers!!!

J: The kind that are going to show up in the GED test?

S: If I was blind, it wouldn’t matter!

J: I’d be showing you in braille or kinesthetically.  Can we just do one with just numbers?  Please please me?

S: Yes.  But, after that, we go back to normal.  Also, Stu is now 5, because he was left out originally.

J: 2/3 ÷ 5/7

S: Um…  2/3 · 7/5 = 14/15.

J: Yes!  Correct!  And that is dividing by fractions!  Shall we do another one?

S: Can this be in sensical terms?

J: Abbreviated, single-letter sensical terms.

S: IE/Y sensical terms?

J: Like this: pj/g ÷ rp/sbj

S: There is no b!  Unless you’re talking about Brian.

J: We need a letter for Pete!  So, b!

S: Ohhhhhh.  Because, otherwise, the variable would clash with Paul.  By the way, you can use regular numbers if they are smaller than -Stu or bigger than Stu.

J: So, let’s simplify our equation.

S: pj/g · sbj/rp

J: And now, we can multiply that!

S: b/24, or just b.

J: Let’s not go down to numbers.  It’s too early!  Please combine this.

S: pjsbj/grp

J: And does anything stand out when you read that numerator?

S: Yes.  I am aware that there are Paul Johns.

J: Well, it depends on what the value of Paul and John are!

S: JOHN IS ALWAYS 1!  PAUL IS ALWAYS 2!  OMG!!!

J: Let’s not let on that we have presumed to value them so.  They might be offended.

S: It’s just MAAAAAATH!!!

J: Second thing: What can be cancelled to simplify the equation?

S: Paul cancels, and John needs to be j^-p.

J: Can we reduce it and then put it in Beatles canonical order?

S: (b · j^p · s)/(gr)

J: Thank you!  Now, feel free to plug in numbers.

S: Pete Best.

J: Yes.  But that’s only because Pete Best’s value was zero.  May I write an equation?

S: Yes, BUT I want it to be like this: Johnny, Paulie, Georgie, Ringsy, Pete, Stu.  No abbreviations!

J: (Johnny + Paulie)/(Georgie + Ringsy) ÷ (Paulie + Pete)/(Johnny + Stu) = …?

S: Georgie/7 · 6/Paulie = 18/14

J: I don’t know.  I was hoping you could solve it in terms of just them.

S: Okay, let’s see… (Stu · Georgie + Georgie)/(Paulie · Stu + Ringsy)

J: That’s great.  I see what you did.  Can we try it another way without substituting traditional numbers?  [(Johnny + Paulie)/(Georgie + Ringsy)] · [(Johnny + Stu)/(Paulie + Pete)]  And for the next step, we could multiply the numerators and denominators, depending on what kind of answer we want.

S: [(Johnny + Paulie) · (Johnny + Stu)]/[(Georgie + Ringsy) · (Paulie + Pete)]     Happy now?

J: Almost.  So, we could use FOIL to figure this out.

S: Alright, what’s the point?

J: The point is, you could do that, and it would be equivalent.

S: Bye!

Reapplying Adding Fractions

March 19th, 2017

S: I forgot how to add fractions.

J: 1/2 + 2/3

S: I’m going to see if I can solve that without any help…  1 is the greatest common factor of 2 and 3…  I have no idea what to do next.

J: It’s the least common multiple, actually.

S: So, 6.

J: Yes, but you didn’t necessarily have to find the least common multiple.

S: Oh yeah?

J: Just recall that anything can be multiplied by 1 and it’s the same number.  So, we can multiply the 1/2 term by 3/3, and it’s still 1/2.  And we can multiply the second term, 2/3, by 2/2, where the 2 is coming from the denominator of the first term.

S: Now I’ll try again.  1/2 + 2/3     1/2 · 3/3 = 3/6     2/3 · 2/2 = 4/6     There.  I knew the denominator would be 6!

J: Cool.  So, now add the two terms together.

S: 3/6 + 4/6 = 7/6, or 1 1/6.

J: Yes, that’s correct.

S: Now, tell me what the story problem behind that was.

J: Well, George eats 1/2 of a peanut butter sandwich before lunch and 2/3 of John’s leftover egg sandwich after lunch.  So, how much sandwich has George eaten?

S: I know!  1 and 1/6 sandwiches!  This time, we can start with a story problem, and I’ll try to solve it without help.  And without watching Help!.

J: Paul has written a melody in 3/4 time.  And John insists that it be scored with his melody that is in 7/3 time.  What will be the time signature of the resulting score?

S: Poor George Martin has to do the math.  If we weren’t specifically adding fractions, I would have no idea what the equation would be.

J: Well, guess what?  It’s not necessarily adding fractions.

S: But that’s the title!

J: I know.

S: I was just thinking that the problem does not make sense with adding.

J: You were?  Great!  How do you think it does make sense, then?

S: Well, first I want to say that time signatures can change in the middle of a song, so it doesn’t matter.  The problem has no meaning.

J: You just want to argue?  Or do you want to figure out what the answer is?

S: Well, you know that I love arguing, but, fine: THE ANSWER.  THE MEANINGLESS ANSWER.

J: Okay.

S: I shall solve it…as soon as I figure out what the equation is.  Multiplication?

J: The trick is to do the same conversion that you do when you’re adding fractions.

S: But, is the full equation multiplication?

J: Yes.  Yes it is.  Just multiply the two fractions and you’ll get your answer.

S: Good.  Multiplication is easy.  3/4 · 7/3 = 21/12, or 1 9/12, or 1 3/4.

J: Mm-hmm.  But the time signature would just be the fractional representation.

S: Wait!  Because this is Beatles math, I must say that “Revolution” just randomly came on the neighbors’ speaker system a few second ago!  Beautiful coincidence.  Beatles math + Beatles song that was not played on purpose.  Anyway, the time signature would be 21/12, if that was how music scoring worked, which it isn’t.

J: So, you could reduce that.

S: To 1 3/4.

J: Let’s do it in terms of a fraction with two integers.

S: So, 7/4.  Hey!  That uses the denominator of 3/4 and the numerator of 7/3!

J: It’s just an accident that it worked out that way.

S: Oh.  Anyway, imagine counting like 1-2-3-4-5-6-7, 1-2-3-4-5-6-7, 1-2-3-4-5-6-7!  Next problem!

J: So, 2/5 of the time, Ringo is practicing with all the Beatles, and 3/7 of the time, Ringo is practicing with either himself or only one or two of the other Beatles.

S: Hang on a minute!  Ringo never practiced by himself because he didn’t need practice because he was the best drummer in the world!

J: But he wasn’t even the best drummer in the Beatles!

S: The point is, he didn’t practice by himself, so the other half of the story problem should just be with one or two other Beatles, not just with himself.

J: But the point of this is to get you to add those two fractions together to find out how much of the time he is practicing.

S: Yes.  Okay.       2/5 + 3/7     2/5 · 7/7 = 14/35     3/7 · 5/5 = 15/35     So, 14/35 + 15/35 = 29/35.

J: Yes.  Correct!

S: So, this means that I can now add fractions without help.  Give me one more problem.

J: This is contentious, but just go with it, alright?

S: Okay.

J: John is 7/5 as good as Elvis, Paul is 8/3 as good as Elvis, George is 3/2 as good as Elvis, and Ringo is 5/2 as good as Elvis.

S: Where is this data coming from?

J: I said it was contentious, so you don’t get an answer.  So, how much better than Elvis is the average Beatle?

S: So, add them together and take the mean of the sum…  Ooh!  George and Ringo have common denominators, so I’ll start with them…  3/2 + 5/2 = 8/2, or 4.  Now for John and Paul.     7/5 + 8/3     7/5 · 3/3 = 21/15     8/3 · 5/5 = 40/155     21/15 + 40/15 = 61/15     4 + 61/15, or 4/1 + 61/15…     41 · 15/15 = 60/15     61/15 · 1/1 = 61/15     60/15 + 61/15 = 181/15     181/15 ÷ 4 (We divide by four because there were originally four fractions.) CALCULATOR…  On average, each Beatle is 3.016 times better than Elvis.

J: Yes.  Very good.

S: The end!!!

Adding Fractions

MARCH 4TH, 2017

J: The Beatles will have pie.  The next task is adding up all the pie.

S: Wait!  We need to add it in a special, more Beatle-y way than just adding.

J: John wants a 1/3 slice of pie.

S: Wow.  That’s a lot of pie.  But George was the hungry one who would ACTUALLY want 1/3 of a pie.

J: George wants all the rest of the pie after the other three have their slices.  Paul wants a 1/5 slice of pie.

S: George is going to get, like, nothing!

J: We’ll see.  Ringo wants a 1/7 slice of pie.

S: For fractions, isn’t there Euclid’s algorithm?

J: Yes.  You’re thinking of the least common multiple, and that’s exactly what you want here.  So, how much pie does George get?

S: Hmm… Well, I don’t remember Euclid’s algorithm!

J: That’s okay!  We don’t have to use it because I cleverly picked prime numbers for the denominators.

S: I still don’t know what to do.

J: Well, George gets the leftover pie, right?

S: Yes.

J: And that would be one pie minus John’s slice, Paul’s slice, and Ringo’s slice.

S: Yes.  But you aren’t supposed to draw a picture when you add fractions.

J: You can if you want to, but we don’t have to.  Let’s write how much John, Paul, and Ringo have all together.

S: That’s what I’ve been trying to do!

J: So, do it!

S: 1/3 + 1/5 + 1/7

J: These have different denominators.  We need to give them-

S: COMMON DENOMINATOR!  COMMON DENOMINATOR!  I remember how to do that!  Wait.  Too bad their only common factor is 1.  But, multiples!  So, 7 · 2 = 14; no; 7 · 3 = 21; no; oh!  I remembered 35, which is 5 · 7, and if I multiply 35 by 3… 35 · 3 = 105.  I got the least common multiple!

J: Yes!  Correct!  So, how many 105ths of a pie does John get?

S: However many times 3 goes into 105?

J: Correct.

S: Then it’s 35/105, or 1/3.  We’re back to where we started.

J: No, no, no!  Keep the 35/105!

S: This is not Beatle-y enough.  We have to find a Beatle-y way to actually DO the math.

J: Can’t we sneak that in at the end?

S: Oh, fine.  I just want this to be over as soon as possible… BECAUSE I’M NOT HAVING FUN!!!

J: So, how much pie does Paul get?

S: 3 · 7 · 1 = 21, so 21/105.

J: Yes.  And-

S: How much pie does Ringo get?  3 · 5 · 1 = 15, so 15/105.  And then we add up the numerators!  15 + 21 + 35 = 71, so, together, John, Paul, and Ringo get 71/105 of the pie.  What flavor is it?

J: It’s pumpkin.

S: If I knew the Beatles’ pie preferences, I could tell you if they’d actually eat it or not, but I don’t because detailed information like that is SO HARD TO FIND just searching Google!!!

J: It’s for a photo shoot.

S: ANOTHER one?

J: Well, how much pie does George get?

S: George gets 34/105.  Less than John by 1/105!  There was no “we’ll see”!!!  I WAS RIGHT!!!

J: But wait!  John gives 1/11 slice of pie to George!

S: Awwwwww…

J: Do you think George has more now?

S: Yes.

J: Because…?

S: 1/105 and 1/11 have the same numerator, and 11 is a smaller number than 105, meaning that it is a bigger fraction.  Wait!  I don’t understand if the 1/11 is 1/11 of the whole pie or 1/11 of John’s slice.

J: The whole pie.

S: Well, then what’s wrong with my reasoning?

J: So, now, this looks fine.  Now, what if John gave George 1/11 of John’s slice of the pie?

S: Ringo is most likely to do that, not John.

J: What if John steps out of the room, and Ringo gives George 1/11 of John’s slice of pie?

S: The point is that Ringo is too nice to do that!  Let’s just have George steal the 1/11.

J: Okay!  1/11 of John’s.

S: So, I should divide 35/105 by 11?  I know how to divide fractions.

J: Well, we could write John’s share in terms of 35/105 · 11/11.

S: That’s what I was going to do!  Except, what does multiplying by 1 do?

J: We’re writing it in terms of the least common multiple!

S: Which isn’t 11!  11 was just a number pulled from your head!

J: So, can you multiply that and find out what John’s share is in terms of those numbers that have 11 as a factor?

S: I understood none of that sentence.

J: Can you perform the mathematical operation indicated seven lines ago?

S: Sure!  35/105 · 11/11 = CALCULATOR.  *gets calculator and types in equation*  Okay.  It equals 385/1155.

J: Ah!  So, what’s 1/11 of that?

S: I know what to do!  On my calculator: whatever fraction is equivalent to 0.030303.

J: I would prefer the answer as a fraction of two integers.

S: Well, I forgot how to convert decimals to fractions, so poo-poo!

J: Well, then let’s go back to this step [five lines ago].  So, in the numerator, we have 35 · 11.  What’s 1/11 of 35 · 11?

S: Oh.  Two minutes left.  And you haven’t done the Beatle-y twist!

J: Well-

S: Ooo!  “Twist and Shout”!  The Beatle-y twist!

J: Well, what’s 1/11 of 385 · 11?

S: Sorry.  Bad reference.

J: Yes, at the end, they will dance to “Twist and Shout” while eating pie.

S: Was that your original plan?

J: No.  But, can you write 35 · 11 / 11?

S: Yes.  It is 35.

J: Yes.  And so, you just divided by 11, but you made it so easy.

S: Now tell me what your original plan was.

J: To solve an algebraic equation of four variables: John, Paul, George, and Ringo.

S: No!  I mean for the Beatle-y twist at the end!

J: I thought it was Beatle-y enough and that it would be done on one page!

Factorial!

FEBRUARY 19TH, 2017

J: John and Paul must stand side-by-side for a concept photo.  How many ways can the photographer arrange them?

S: 2.

J: What are those ways?

S: John on the left and Paul on the right, and Paul on the left and John on the right.

J: George joins them.

S: Now how many ways?

J: NOW how many ways?

S: 6.

J: Ringo shows up.  Now how many?

S: 24.

S: When there were 2 Beatles, there were 2 ways; when there were 3 Beatles, there were 6 ways; and when there were 4 Beatles, there were 24 ways.  2 · 3 = 6, and 6 · 4 = 24!  A PATTERN!!!

J: So, Pete Best wanders by.

S: But if Ringo’s there, he’s already been kicked out.

J: Yes, but these nice guys will let him be in the photo.

S: They won’t.

J: They’re getting paid £10,000,000 to do it.

S: Okay, fine.

J: So, how many Beatles are there?

S: 4, because Pete isn’t a Beatle anymore.

J: How many Beatles, new and old, are there?

S: 5.

J: And how many ways can the photographer arrange them?

S: 24 · 5 = 120.  So, the photographer can arrange them 120 different ways.

J: How many places are there where Pete could stand?

S: 5.

J: And, for each place Pete stands, how many ways can the other four Beatles be arranged?

S: 24.

J: So, that’s how you get the 24 · 5 = 120.  This 120 is 5 · 4 · 3 · 2.  This thing has a name: factorial.

S: FAAAAAAAAAAAAACTOOOOOOOOORIIIIIIIIIIIIIIIIIAAAAAAAAAAAAAAAAAAL!!!

J: So, we could write it as 5! (five followed by an exclamation point).

S: So, the exclamation point is also a math symbol?

J: Mm-hmm.  Yes.

S: That’s weird.

J: Mm-hmm.

S: Okay.  Go on.

J: Stu shows up.

S: YAAAAAY!  STU!!!

J: How many ways can the photographer arrange them now?

S: 2 · 3 · 4 · 5 · 6, or 6!, otherwise known as 120 · 6, otherwise known as 720.  720 DIFFERENT POSSIBLE PHOTOS.

J: For just 6 people!

S: People! Get it?

J: Hahahaha!

S: People factorial!!!  That’s what you said.  Wait – is this the end of the lesson?

J: Well, what if Brian and Linda show up, AND the other George – what’s his name?

S: George MARTIN!!!  How could you not know?!

J: So, how many people must be arranged in the photo?

S: This doesn’t work.  Stu died long before Paul met Linda!

J: She just happened to wander by!

S: But they didn’t know her!

J: But they needed a woman in the photo!

S: But she lived in New York!

J: And they were doing the photo shoot in New York!

S: It still isn’t realistic.

J: Then who?  Who’s their most famous fangirl?

S: There is no “most famous” fangirl!

J: How many hung out at Apple?

S: I don’t know – a lot – but this is in New York, remember?

J: Weren’t there four that were in the book – George wrote a thank-you to one of them – something like that?

S: Yes, but they’re in New York!  How about Frieda Kelly?  She was their secretary and she started out as a Beatles fangirl in the Cavern days.

J: Okay.  So, how many people must be arranged in the photo?

S: Wait – what  what about Mal and and Neil?

J: Okay.  They get in the photo, too.

S: Now, there are 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 · 11 different ways for the photo to be arranged.  I’m getting a calculator.  Okay.  There are 39,916,800 ways, or 11!.  That photo shoot would take years, and everyone would start fighting with each other.

J: Now, all thousand school chums show up and join them.

S: Like Ivan?

J: Mm-hmm.

S: Do you even know who Ivan is?

J: I remember the name.  And surely, out of one thousand school chums, they had at least one named Ivan.

S: IVAN WAS THE MOST IMPORTANT PERSON IN BEATLE HISTORY!  HE WAS THE ONE WHO INTRODUCED JOHN AND PAUL!!!  So, without him, we wouldn’t be doing Beatles math right now.

J: Oh.  So of course he has to be in the photo!  So, how many people will be in the photo?

S: 1,009.

J: And how many ways must they be arranged?

S: Um…um…something factorial!

J: Yes.  What number factorial?

S: Well, not 1,009, because when there were 9 people, it was 11!

J: No!  I thought there were 11 people for 11 factorial!  No, no, no…  There’s been a mistake.  I thought we were going from 11 people to 1,000!  So, I don’t know where the 9 came from…

S: Let’s review it…  Oh!  It WAS 11!  I forgot about Mal and Neil!  So, 1,011! (or one thousand eleven factorial).  NOW are we done?

J: Not yet.

S: NOT YET?

J: No, but I won’t make you type in 1,011 multiplications on a calculator.  So, I want to say something: For a lot of school calculators, 69! is the highest the calculator can handle.

S: And on mine, it doesn’t even have factorial!  Okay, I looked up 1,011! on your phone calculator, and it is 4.297327055E2600.  NOT A REAL NUMBER!  I LOATHE WHEN CALCULATORS GIVE “E” AS AN ANSWER!!!

J: It actually is a real number, as opposed to a natural number.  Oh, I misspoke.  So, it’s clearly kind of fuzzy, right?  But it’s not very satisfying, is it?

S: It’s not.  Which is why I loathe “E”.

J: That’s a pretty big number.

S: Bigger than undecillion?

J: Yes.

S: WOW.

J: So, let’s say that all the people are fast, and they can rearrange themselves for each photo in just one second.  So, how long will it take to take all the photos?

S: 4.297327055E2600 seconds.

J: Uh… How many minutes?

S: 4.297327055E2600 ÷ 60.

J: Well, there are 86,400 seconds in a day.  And 365 days in a year.

S: Let me do 86,400 · 365… Oh!  It’s only 31,536,000!  Uh-oh.  That means they’ll be taking photos for a LOOOOOOOOONG time.  What are you doing with my calculator?

J: Just a silly little exercise.  So, how many years will it take to take all the photos?

S: Another dumb “E”!  1.355205060376327403635…

J: Wait, you multiplied!  We want to divide by that.  We divide by the number of seconds per year to find out how many years.

S: Another dumb “E”!  What’s the difference?

J: Indeed.  How old is the universe?

S: Around 13,000,000,000 years old.

J: Sure.  So, is it practical to take photos of all those arrangements?

S: No.

J: Correct.

S: Is that the end of the lesson?

J: Tiny little wrap-up: For 4 Beatles, we had 4! arrangements.  For 3 Beatles, we had 3! arrangements.  When there were just John and Paul, we had 2! arrangements.

S: But not even 2!!  It was just plain old 2.  Please tell me we’re done.

J: Almost.  So, can you picture the different arrangements – the different ways to arrange John and Paul?

S: Yes.

J: So that’s HOW many ways?  2 factorial.  2! ways to arrange them.

S: That’s a nice ending, isn’t it?  HINT HINT

J: Almost.  Now, there are two more photos to take.  John is there.  So, how many Beatles are there in the photo?

S: 1.

J: So, how many ways can that one Beatle be arranged?

S: 1. Or 1!.

J: And how much is 1 factorial?

S: 1.

J: John has grown bored of this, and wanders off to catch up with Paul.  How many Beatles are there to arrange?

S: 0.

J: How many ways can the photographer arrange the Beatles?

S: 0!  We’re done!

J: Almost.  So, the photo is taken, so that’s an arrangement.  The photo is taken with 0 Beatles.  So, if we’re counting, how many ways can we arrange 0 Beatles?

S: 0 factorial.

J: How many photos do we have of 0 Beatles?

S: 1.

J: So, 0! is equal to…?

S: 1.  NOT 0.  That’s weird.

J: That’s a special case for the convenience of mathematics working out for us.

S: LESSON CLOSED.

J: Good job!

S: THE END.

Probability

February 12th, 2017

J/S: The Beatles are due on stage in one hour.

Ringo is in the dressing room at the stadium.  It takes him 1/2 hour to choose red things to wear.  It takes him 3 times as long to choose blue things as red.

George is hungry, and there is a 90% chance that he will see a restaurant, a 70% chance that he will go in, and an 80% chance that someone will see him, and a 25% chance that they will know everything about him, and a 60% chance that they will want to study his teeth to see if he really is a vampire for more than an hour.

John is on a date, and dates take 40 minutes per girl, and 3/5 of his dates are with one girl.  40% are with 2 or more girls.

Paul is walking to the stadium, disguised.  Along the way, there is a 30% chance that he will notice George in a restaurant and want to join him, and there is a 15% chance that the George fangirls examining George’s vampire teeth are actually Paul fangirls and they will turn around and kidnap him, and they will have replaced the restaurant employees with robots, and they will have shut down the telephone, and they will have AK-47s, so George won’t be able to do anything, and there is a 30% chance that the AK-47s will jam, and George will succeed in rescuing Paul.  Back on the street, if none of that happened, there is a 50% chance that Paul will be recognized by fans and chased all the way to the stadium, and will get there early.  He will have nothing to do, so he’ll go to the dressing room and find Ringo there, and they will have a long argument about the colors Ringo should wear, and there is a 50% chance that it will take too long.

S: Here comes the math…  WARNING: This probably won’t make sense. It certainly doesn’t make sense to me.

J: R_{on time} = 50% (1-25%) = 37.5%

PG_OT = 92%, P_late = 88.85%

PG_L = 0.9 · 0.7 · 0.8 · 0.25 · 0.6

PR_OT = 1/2

PG_OT = 1 – PG_L – 1 – 0.0756 = 0.92435

PJ_OT = 60% (40% · 10% PG goes in) – 60% + 4% = 64%

P_LKN = 0.3 · [(G_L = 0.9 · 0.7 · 0.8 · 0.25 · 0.60) = 92%) · 0.15 · (1 – 30%)] = 0.3 · 0.92 · 0.15 · 0.7 = 0.02898

P_LDRA = 50% · 50% = 25% = 0.25

P_L = P_LKN + P_LDRA = 0.02898 + 0.25 = 0.27898

P_OT = (1 – P_L) = 0.72102

J/S: Wilfrid Brambell is going out to lunch.  There is a 10% chance that he will eat at the restaurant where John is and make sure that he won’t be late to the show.  There is a 10% chance that he will eat at Paul and George’s restaurant and decide to be a king mixer there and order 100-year-old scotch that is very expensive, and there is a 50% chance that he won’t get away with it and Paul and George have to rescue him.  There is an 80% chance that he’ll eat somewhere else.

J: Math adjustment: P_OT = 0.72102 – P_late-grandad = 0.71157

10% Grandpa goes in

90% · 70% George goes in

50% rescue needed, George is late

PG_{late-grandad-rescue} = 10% · 90% · 70% · 50% = 0.0315

10% Grandpa goes in

30% Paul not as George goes in · 90% · 70%

50% rescue failed, Paul is late

Paul late Grandad rescue = 0.00945

J/S: Q1: What is the probability that at least one Beatle is on stage on time?

A1: 99.264012375%

Q2: What is the probability that ALL the Fab Four are on time?

A2: 15%

Q3: If it costs £100,000 if a concert must be canceled, and Brian wants to insure their concerts, how much should the insurance company charge per concert at minimum to break even?

A3: £85,000

Equality Play

February 12th, 2017

J: To start off, why don’t we take turns asking each other questions, with each question using something from the previous answer.  You go first.  Ask me a question.

S: Any question?

J: Any question.

S: Okay!  On what day was Pete Best fired from the Beatles?

J: January 1st, 1961.

S: Wrong.  August 16th, 1962.

J: Rewrite the year as a product.

S: 981 · 2

S: Use short division to find 981 ÷ 2.

J: *uses short division to find that the answer is 490.5*

J: Rewrite 490.5 as the sum of four terms.

S: 490 + 0.1 + 0.2 + 0.2

S: How much of a Beatle is 0.2%?

J: *does a bunch of calculations* Paul’s ring fingers (based on estimated weight), 160 grams, with ring.

J: Rewrite 160 grams as a product of primes.

S: 2 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5 · 5

J: 2 · 80 = 2 · 2 · 40 = 2 · 2 · 2 · 20 = 2 · 2 · 2 · 2 · 10 = 2 · 2 · 2 · 2 · 2 · 5 = 2⁵ · 5

Note: This one is a bit weird and short and, well, boring.  Don’t worry; the next one and all the ones following it are much better!  This is just here because it’s the first math involving the Beatles that we did…