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PtichkaEL [24]
3 years ago
13

Is s the square root 6 rational

Mathematics
1 answer:
bogdanovich [222]3 years ago
7 0

Proof by contradiction.

Let assume \sqrt6 is rational. Therefore, it can be expressed as a fraction \dfrac{a}{b}, where \text{gcd}(a,b)=1

\sqrt6=\dfrac{a}{b}\\\\
6=\dfrac{a^2}{b^2}\\\\
a^2=6b^2

This means that a^2 must be even, and therefore also a must be even.

So, a=2k, k\in\mathbb{Z}

(2k)^2=6b^2\\\\
4k^2=6b^2\\\\
2k^2=3b^2

This means that 3b^2 must be even. The only way for this to be even is when b^2 is even and therefore also b is even.

But if a and b were even, then earlier assumption that \text{gcd}(a,b)=1 would be false. Therefore \sqrt6 is irrational.

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