The Mathematical Proof Thread

[Kernel Sohcahtoa]

Via Lara Alcock (How to Study as a Math Major, 2013: oxford university press), math proofs are like an internal combustion engine: we drive just fine without knowing how a car actually works, but if we are ever curious why a car operates the way it does, then we need to understand the internal combustion engine, which is the broader system that makes driving a reality. 

I've created this thread for people to post, share, discuss, or inquire about any proof they want.  It would be really cool if the proof was something beautiful to you. Perhaps we could even form a mini proof library.  

Here's a basic example to get things rolling.

Inquiry: suppose we have the even numbers 2,4,6.  We know that these numbers are even.  We also know that if we square these numbers then the squares are also even.  But, although we intuitively know this, how could we show that this result is true for 2,4,6, or more importantly, for all even integers x?

Definition of an Even Number: An integer n is even if n=2a for some integer a. (Hammack, Book of Proof, 2013, pg 89)


Proposition: If x is an even integer, then x^2 is even. 

Proof (direct). Suppose x is an even integer.  Then x=2a for some integer a via the definition of an even number.  Now substitute x=2a into x^2, which gives x^2=(2a)^2=4a^2=2(2a^2).  Consequently, x^2=2b for some integer b=2a^2.  Thus, x^2 is even by the definition of an even number.

[Excited Penguin]

Don't know much math, but that sounds like WIFOM to me.

[Joods]

My kid is studying some stuff in her 8th grade advanced honors algebra class. I'll have to run this by her to see if she understands. Once you start throwing letters into a math equation, I get lost. Letters don't belong with numbers for us dumb folk.

[TheRealJoeFish]

I do love me some proofs.

I remember for whatever reason the proof of Borel's Lemma being particularly interesting to me back in Advanced Number Theory but that could just be me remembering the joy of finally getting to any sort of measurable accomplishment - the *entire* semester was spent proving the Cayley-Bacharach Theorem. It's been nearly 5 years since I've done any higher-level mathematics like that with any consistency, so both of those proofs are currently wayyy beyond my ability to explain.

P.S. I'll post some cool proofs that I *do* remember and understand soon!

[GrandizerII]

My kid is studying some stuff in her 8th grade advanced honors algebra class. I'll have to run this by her to see if she understands. Once you start throwing letters into a math equation, I get lost. Letters don't belong with numbers for us dumb folk.

if she's doing advanced honors algebra, then this is just basic stuff for her.

[Whateverist]

This wasn't presented as an occasion for producing a proof but I found it a fun problem anyway.  I wonder what you think of my argument in favor of the proposition.  (I'll hide it in case you want to have a go at it yourself first.)

Proposition:  The cube of any odd number ≥ 3 decreased by that same odd number will always be divisible by 24

---

The prime factorization of 24 is 2^3 • 3.  So to show n^3 – n (where n is an odd number ≥ 3) must be divisible by 24, we must show it has at least three factors of 2 and at least one factor of 3.

It is easy enough to show that n^3 – n must have three factors of 2,

since n^3 – n = n(n^2 – 1) = n(n – 1)(n + 1) = (n – 1)n(n + 1)

If n is odd, then it can be written as 2a – 1 and
then (n – 1) = 2a – 1 – 1 = 2a – 2 = 2(a – 1), an even number
and (n + 1) = 2a – 1 + 1 = 2a, also an even number
In fact, 2a and 2(a – 1) are consecutive even numbers and that means one of them is also a multiple of 4, or 2^2
Thus n^3 – n has at least 3 factors of 2 and so is divisible by 2^3.

But why must it have a factor of 3?  In the first allowable case, where n = 3, n itself is a multiple of 3.  
However, in the very next case, where n = 5, n lacks a fact of 3.  In this case, n – 1 = 4, and, n + 1 = 6.  So it is 6 which carries the factor of three.  Since every third counting number is a multiple of three, and since (n – 1)n(n + 1) is the product of three consecutive counting numbers, (n – 1)n(n + 1) will always contain exactly one factor of three.

Since n^3 – n must always contain at least 3 factors of 2 and exactly 1 factor of 3, the cube of any odd number (3 or larger) decreased by that same odd number will always be divisible by 3.

---

[Alex K]

Whatevers, cool, after a first reading, it looks watertight to me. As far as I am concerned, you could shorten your second step to
"
n^3-n=(n-1) * n * (n+1)
If n is odd, n-1 and n+1 are even. In fact, one of them is divisible by four because that os always the case for two consec. even numbers.
"
without introducing a. That made it harder to read and I don't see why you needed it.

[Alex K]

I'll try to write an induction proof for it later, the problem seems perfectly suited for that

[Whateverist]

Whatevers, cool, after a first reading, it looks watertight to me. As far as I am concerned, you could shorten your second step to

n^3-n=(n-1) * n * (n+1)
If n is odd, n-1 and n+1 are even

without introducing a. That made it harder to read and I don't see why you needed it.

Glad to hear you say so.  Felt like sticking legs on a snake to me too.  I guess I was trying to make it more math-y.

[Alex K]

The more I think about it, the more it becomes apparent that an induction proof could only be more complicated than yours because it would basically contain it.