Question

prove that 3/7​ root 5 is also an irrational number

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Answer 1

Answer:

Let us assume that √5 is rational

So,

√5 = a/b. where a and b are integers such that they're coprime to each other.

So,

a^2 = 5b^2

So,

a^2/5 = b^2

which means 5 divides p^2

which indirectly also means, 5 divides p

Hence, we can say

a/5 = c where c is some other integer

So,

a = 5c

a^2 = 25c^2

5b^2 = 25c^2

b^2/5 = c^2

which means 5 divides b^2 and again indirectly that 5 divides b

This means, 5 divides both a and b

Hence, 5 is a factor of both a and b.

Therefore, a and b are not coprime

This contradicts our assumption and thus our assumption is false

Therefore, √5 is irrational number.

Now,

Let us assume that 7-3√5 is rational.

So,

7-3√5 = p/q where p and q are integers co prime to each other.

So,

(7-p/q) = 3√5

So,

√5 = (7-p/q)/3

Now, here at the Left hand side we have an irrational number.

and on the right hand side we have a rational number because even p and q are integers.

This contradicts the equality

and hence our assumption is incorrect , and ultimately hence, 7-3√5 is also an irrational number

Hope this helps you!