Square Root of 2 is Irrational/Classic Proof

Theorem

2 √ is irrational.

Proof

First we note that, from Parity of Integer equals Parity of its Square, if an integer is even, its square root, if an integer, is also even.
Thus it follows that:

(1):2∖p 2 ⟹2∖p

where 2∖p indicates that 2 is a divisor of p .
Now, assume that 2 √ is rational.
So 2 √ =p q for some p,q∈Z and gcd(p,q)=1 .
Squaring both sides yields:

2=p 2 q 2 ⟺p 2 =2q 2

Therefore from (1) :

2∖p 2 ⟹2∖p

That is, p is an even integer.
So p=2k for some k∈Z .

Thus:

2q 2 =p 2 =(2k) 2 =4k 2 ⟹q 2 =2k 2

so by the same reasoning:

2∖q 2 ⟹2∖q

This contradicts our assumption that gcd(p,q)=1 , since 2∖p,q .
Therefore, from Proof by Contradiction, 2 √ cannot be rational.
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Historical Note

This proof is attributed to Pythagoras of Samos, or to a student of his.
The ancient Greeks prior to Pythagoras, following Eudoxus of Cnidus, believed that irrational numbers did not exist in the real world.
However, from the Pythagorean Theorem, a square with sides of length 1 has a diagonal of length 2 √ .
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