Question

Proof by contradiction: Suppose a right triangle has side lengths a, b, c that are natural numbers. Prove that at least one of a, b, or c must be even. (Hint: Use Pythagorean Theorem)

Answer 1

Suppose, a is the hypotenuse.

Then by Pythagoras theorem,

a² = b² + c²

Suppose, none of a, b, c is even.

i.e. all of them are odd, so, a, b & c are odd all.

Then, the squares of odd numbers is also odd.

So, a², b² & c² are all odd.

Now, a² = b² + c² = odd + odd = even

So, a² = even, which implies a is even.

This is a contradiction that a is odd (along with b & c)

So, not all of a, b & c are odd.

So, at least one of a, b & c must be even.