Question

Prove that root 7 is irrational by the method of contradiction

Answer 1

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

`\sqrt7=(a)/(b)\\\\7=(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.