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Answer 1

Q. Show that \sqrt{7} is a irrational number.

TO PROVE: 2/√7 is an irrational number…

For proving the above, firsthand we prove the fact that √7,…. & in general √p is an irrational number, where p is a prime.. ( just like the proof of, √2 is an irrational number)

Using this fact , we can now prove the required by contradictory method as follows :

2/√7 * √7/√7 ( rationalised the denominator)

= 2√7 / 7

Now, let us assume that 2√7 / 7 is a rational number

=> 2√7/7 = p/q ( where, p & q are integers, q not equal to 0)

=> 2√7 = 7p/q

=> √7 = 7p/2q

Here, LHS is an irrational number( as it has been already proved)

Whereas RHS is a rational number, because in 7p/2q, 7p is an integer, also 2q is an integer, as product of 2 integers= integer. & 2q is not equal to 0 ( as we have mentioned above that q is not equal to 0). This way all the conditions of a rational number are satisfied…..

So, LHS is not equal to RHS. As one side is rational & the other side is irrational…

This contradiction arises because of our wrong assumption.

This proves that 2√7/7 OR 2/√7 is an irrational number.

Answer 2

Answer:

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