Question

Que. 10 Prove that (5 + unferroot2) is an irrational number.
give me answer fast plz and briefly proof​

Answer 1

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Answer 2

Answer:

To prove : 5 + √2 is an irrational number.

Proof :

Let is assume that √2 is a rational number. Then ,

√2 = p/q

{ where p and q are co-prime numbers and q is not equal to 0 .}

√2 = p/q

Squaring both sides, we get

(√2)^2 = (p/q)^2

=> 2 = p^2/q^2

=> 2q^2 = p^2

=> p^2 = 2q^2 ...(1)

Here, 2 is a factor of p^2.

Hence, 2 is a factor of p.

Now,

p = 2m ...(2)

( where m is any non-negative integer.)

Applying the value of p = 2m from eq.(2) to eq.(1) , we get

(2m)^2 = 2q^2

=> 4m^2 = 2q^2

=> 2m^2 = q^2

=> q^2 = 2m^2

Here, 2 is also the factor of q^2.

Hence, 2 is also the factor of q.

From above results, we get that

2 is the factor of both p and q.

But this is not possible because p and q are co-prime numbers.

This is a contradiction to our assumption that √2 is a rational number.

Hence our assumption is wrong.

So √2 is an irrational number.

Now let us assume that ( 5 + √2 ) is a rational.

Then,

( 5 + √2 ) = p/q

{ where p and q are co-primes and q is not equal to 0. }

5 + √2 = p/q

=> √2 = p/q - 5

=> √2 = ( p - 5q ) /q

Here, {( p - 5q ) / q} is a rational number but √2 is an irrational number ( proved above ).

So here, L.H.S. is not equal to R.H.S.

This is a contradiction to our assumption that ( 5 + √2 ) is a rational number.

Hence our assumption is wrong.

Thus,

( 5 + √2 ) is an irrational number.

Hence proved.