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Ulleksa [173]
4 years ago
5

Prove that root 7 is irrational by the method of contradiction

Mathematics
1 answer:
Alchen [17]4 years ago
8 0

Let assume that \sqrt7 is a rational number. Therefore it can be expressed as a fraction \dfrac{a}{b} wherea,b\in\mathbb{Z} and \text{gcd}(a,b)=1.

\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2

This means that a^2 is divisible by 7, and therefore also a is divisible by 7.

So, a=7k where k\in\mathbb{Z}

(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2

Analogically to a^2=7b^2 ------- b^2 is divisible by 7 and therefore so is b.

But if both numbers a and b are divisible by 7, then \text{gcd}(a,b)=7 which contradicts our earlier assumption that \text{gcd}(a,b)=1.

Therefore \sqrt7 is an irrational number.

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