Sqrt(2) is irrational

Definitions:

  1.  An integer is called even if and only if it is divisible by 2, that is some integer \displaystyle r is even if for some other integer \displaystyle s the statement \displaystyle r = 2s is true.
  2. An integer is called odd if and only if it is not divisible by 2, that is some integer \displaystyle n is odd if for some other integer \displaystyle m the statement \displaystyle n = 2m + 1 is true.
  3. A number is called rational if it is in the form of \displaystyle \frac{p}{q} where \displaystyle p and \displaystyle q are integers. Further, it is said to be in lowest terms if \displaystyle p and \displaystyle q are not commonly divisible by any integer save the identity 1. A number that is not rational is called irrational.
  4. A number \displaystyle x is called a square root if it satisfies \displaystyle x^2 = y for some number \displaystyle y where for our purposes \displaystyle y is a positive integer. “The square root of \displaystyle y” is denoted by \displaystyle \sqrt{y}.

Theorem 1 The product of two even integers is even.
Proof For any two even integers \displaystyle r = 2k and \displaystyle s = 2j, we have

\displaystyle rs = (2k)(2j) = 2(2kj)

which is even. QED.

Theorem 2 The product of two odd integers is odd.
Proof For any two even integers \displaystyle p = 2m + 1 and \displaystyle q = 2n + 1, we have

\displaystyle pq = (2m + 1)(2n + 1) = 4mn + 2m + 2n +1 = 2(2mn + m + n) + 1

which is odd. QED.

Corollary 1 If the square of a number is even, then the number itself is even. If the square of the number is odd, the number is likewise odd.

Theorem 3 \displaystyle \sqrt{2} is irrational.
Proof Assume that \displaystyle \sqrt{2} is rational and in lowest terms, such that

\displaystyle 2 = \frac{p^2}{q^2}

or

\displaystyle p^2 = 2q^2

that is \displaystyle p^2 is even. Then, by Corollary 1, \displaystyle p is even as well. That is,

\displaystyle p = 2n

for some integer \displaystyle n. Therefore,

\displaystyle (2n)^2 = 4n^2 = 2q^2

which implies

\displaystyle q^2 = 2n^2

and thus \displaystyle q^2 is even and therefore \displaystyle q is even, again by Corollary 1. Therefore, both \displaystyle p and \displaystyle q are even which contradicts our initial assumption that \displaystyle \sqrt{2} was rational. Therefore, the assumption was false and \displaystyle \sqrt{2} is irrational via proof by contradiction. QED.

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