Question

Prove that 7√5 is irrational number​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Let us assume the opposite in order to prove that it is irrational.

Let

$7 \sqrt{5} \: \: \: be \: \: \: rational \: \: \: number.$

So, it can be written in the form of p/q where p and q are co - prime numbers and q is not equal to 0.

Hence,

$7 \sqrt{5} = \frac{p}{q} \\ = > \sqrt{5} = \frac{1}{7} \times \frac{p}{q} \\ = > \sqrt{5} = \frac{p}{7q}$

$here \: \: \: \sqrt{5} \: \: is \: \: irrational \: \: and \: \frac{p}{7q} \: \: is \: \: irrational.$

Since, rational is not equal to irrational.

This is a contradiction.

Hence, our assumption is wrong.

Hence, proved.