Question

prove that 5-√7 is irrational​

Answer 1

Answer:

Let us assume that √7 is rational

√7 =a/b, where a and b are co prime

√7b=a

squaring both sides

7b^2=a^2            --------(1)

7 divides a^2, therefore 7 divides a ( theorem 1.3, let p be a prime number. If                         p divides a^2 then p divides a, where a is a positive integer)

We can write a=7c, for some integer c

substituting value of a in (1),

7b^2=(7c)^2

7b^2=49c^2

b^2=7c^2

7 divides b^2, therefore 7 divides b (theorem 1.3, let p be a prime number. If p divides a^2 then p divides a, where a is a positive integer)

Therefore a and b have at least 7 as common factor

but this contradicts the fact that a and b are co prime

This contradiction has arisen because our assumption is wrong

thus √7  is irrational

Let us assume 5-√7 is rational

5-√7 = a/b, where a and b are co prime

√7=5-a/b

LHS is irrational, RHS is rational

This contradicts the fact that √7 is irrational

This contradiction has arisen because our assumption is wrong

Thus 5 - √7 is irrational

Step-by-step explanation:

You will have to prove that √7 is irrational because it is not given in the question

Hope this helps :)