Question

without using division method show that under root 7 is an irrational no.​

Answer 1

Answer:

Here 7 divides b² so for 7 divides b. Here we find 7 is common which divide both a and b but this is contradiction because a and b are co prime they don't have common factor other than 1. So for our assumption is wrong. Hence √7 is irrational.

Answer 2

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Question:-

• Show that √7 is a irrational number

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Answer:-

Let assume that √7 is a rational number

So it's can be equal to P/q

$> \sqrt{7} = \frac{p}{q} \\ \\ > 7 \: = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ > 7 {q}^{2} = {p}^{2}$

By using Fundamental theorem of arithmetic

$> p \: = 7r \: for \: some \: integer \: r$

Now, square both side

$> {p}^{2} = 49 {r}^{2} \\ \\ > 7 {q}^{2} = 49 {r}^{2} \\ \\ > {q}^{2} = 7 {r}^{2} \\ \\ > 7 | {q}^{2} \\ > 7| q$

So our contradiction is wrong √7 is a irrational number.

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Value of √7 is 2. 645751....

As we know irrational number divide can never be end do √7 is 2.645751 is not a recurring number nor terminating and it's divide can never be end . so it's a rational number.

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