Prove that no isosceles right triangle exists whose sides are integers.
let the shorter sides be x units each ,
let the hypotenuse be h
(clearly we can't have the equal sides being the hyppotenuses, or else we would have 2 right angles, leaving nothing left for the third angle)
h^2 = x^2 + x^2
h^2 = 2x^2
h = √2x
so whatever integer x is, multiplying an integer by √ makes it irrational, thus no longer an integer.
To prove that no isosceles right triangle exists whose sides are integers, we need to use a proof by contradiction.
Assume for contradiction that there exists an isosceles right triangle with integer sides. Let's denote the length of the two congruent sides as 'a' and the length of the hypotenuse as 'c'. Since it's a right triangle, we can apply the Pythagorean theorem:
a^2 + a^2 = c^2
2a^2 = c^2.
From this equation, we can infer that 'c' must be an even number since it is the square of 'a', which is an integer. Let's say c = 2m for some positive integer m.
Substituting the value of c, we get:
2a^2 = (2m)^2
2a^2 = 4m^2
a^2 = 2m^2.
Now let's analyze the equation a^2 = 2m^2. If a^2 is divisible by 2, then a must also be even because for a number to be divisible by 2, it has to be even. So, let's say a = 2n for some positive integer n.
Substituting the value of a, we get:
(2n)^2 = 2m^2
4n^2 = 2m^2
2n^2 = m^2.
From this equation, we can see that m^2 is divisible by 2, which implies that m must be even. Let's say m = 2k for some positive integer k.
Substituting the value of m, we get:
2n^2 = (2k)^2
2n^2 = 4k^2
n^2 = 2k^2.
This implies that n^2 is divisible by 2, which in turn means n must also be even. Therefore, we have found that both a and n are even, but this contradicts our initial assumption that a is an integer.
Hence, our assumption that there exists an isosceles right triangle with integer sides leads to a contradiction. Therefore, we can conclude that there are no isosceles right triangles whose sides are integers.