Question

prove that 2/sqrt(7) is irrational

Answer 1

To prove 2/root 7 is irrational, you have to show that root 7 is irrational, as you know that when a rational number is divided by irrational no. then the no. obtained is irrational.

To prove- Root 7 is irrational.

Proof-

We can do this by method of contradiction.

Let root 7 is rational.

=> root 7 =p/q, where p and q are coprimes and q is not equal to 0.

=> 7=(p^2)/(q^2)

=>p^2= 7q^2

=>7 is factor of p^2

=> 7 is factor of p

=>p=7k, where k is a constant

=>p^2=49k^2

=>7q^2=49k^2

=>q^2=7k^2

=> 7 is also a factor of q^2 and thus a factor of q.

p and q have 7 as a common factor except 1.

This is a pure contradiction to the fact that p and q are coprimes i.e they have only 1 as the common factor.

So our assumption that root 7 was rational is wrong.

Root 7 is irrational.

So 2, a rational no. When divided by root 7 the answer is irrational.

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Answer 2

If possible, Let 2√7 be a rational number.
then, 2√7=p/q (where p and q are co primes)
√7=p/2q
Now, let √7 be a rational no.
then, √7=p/q(where p and q are co primes)
√7q=p
squaring both sides
(√7q)^2=(p) ^2
7q^2=p^2
p^2=7q^2+0 (compare it with a=bq+r)---------1
7 is factor of p^2
so, 7 is also factor of p---------2
By division lemma
p=7m
put p=7m in 1
(7m) ^2=7q^2
49m^2=7q^2
7m^2=q^2
7 is a factor of q^2
so, 7 is a factor of q---------3
From 2 and 3
7 is a factor of both p and q
therefore, our assumption was wrong.
therefore, √7is irrational.

So, √7 is irrational
therefor 2√7 is also irrational.
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