Why is the square root of 5 an irrational number?

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Why is the square root of 5 an irrational number?

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Answer 1 · 2016-03-30T06:41:35

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Explanation:

Here's a sketch of a proof by contradiction:

Suppose $sqrt(5) = p/q$ for some positive integers $p$ and $q$ .

Without loss of generality, we may suppose that $p, q$ are the smallest such numbers.

Then by definition:

$5 = (p/q)^2 = p^2/q^2$

Multiply both ends by $q^2$ to get:

$5 q^2 = p^2$

So $p^2$ is divisible by $5$ .

Then since $5$ is prime, $p$ must be divisible by $5$ too.

So $p = 5m$ for some positive integer $m$ .

So we have:

$5 q^2 = p^2 = (5m)^2 = 5*5*m^2$

Divide both ends by $5$ to get:

$q^2 = 5 m^2$

Divide both ends by $m^2$ to get:

$5 = q^2/m^2 = (q/m)^2$

So $sqrt(5) = q/m$

Now $p > q > m$ , so $q, m$ is a smaller pair of integers whose quotient is $sqrt(5)$ , contradicting our hypothesis.

So our hypothesis that $sqrt(5)$ can be represented by $p/q$ for some integers $p$ and $q$ is false. That is, $sqrt(5)$ is not rational. That is, $sqrt(5)$ is irrational.

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Why is the square root of 5 an irrational number?

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