Prove that √8 is irrational
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you would go with a proof by contradiction:
suppose √8 = a/b with integers a, b
and gcd(a,b) = 1 (meaning the ratio is simplified)
then 8 = a²/b²
and 8b² = a²
this implies 8 divides a² which also means 8 divides a.
so there exists a p within the integers such that:
a = 8p
and thus,
√8 = 8p/b
which implies
8 = 64p²/b²
which is:
1/8 = p²/b²
or:
b²/p² = 8
which implies
b² = 8p²
which implies 8 divides b² which means 8 divides b.
8 divides a, and 8 divides b, which is a contradiction because gcd (a, b) = 1
therefore, the square root of 8 is irrational.
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