Question

prove that 7-root 7 is irrational

Answer 1

Suppose 7-√7 is rational.Then,there exists co-prime such as,
7-√7 = a/ b
=> √7 = 7- a/b
=> √7 = (7b-a)/b
Now, we know that √7 is irrational no. but here (7b-a)/b comes a rational no. and this is a contradiction.
So, our supposition is wrong.
Hence, 7-√7 is an irrational no.
Proved....


Hope it helps....

Answer 2

Let us assume that 7-√7 is a rational number.

7-√7=p/q (where p and q are rational and q is not equal to zero.)

-√7=p/q -7

-√7=p-7q/q

√7=-(p-7q) /q

√7=7q-p/q... eqnA.

Let us assume that√7 is rational.

√7=p/q( where p and q are Co primes and q is not equal to zero).

Squaring on both sides.

7=p^2/q^2

p^2=7q^2...eqn1.

7 divides p^2

Therefore, 7 divides p
Hence 7is factor of p.

p/7=c

Squaring on both sides.

p^2/49=c^2

p^2=49c^2....eqn 2.

From eqn 1 and 2.

7q^2=49c^2

q^2=7c^2

7divides q^2.

Therefore 7 divides q.
Hence 7is factor of q.

But p and q are Co primes.
Therefore this contradicts our assumption.
Thus, we can say that √7 is irrational.

From eqn A
Irrational is not equal to rational.
LHS is not equal to RHS.
Thus, 7-√7 is an irrational number.

Hence proved.