Question

Prove that 5+3√2 is an irrational number

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Let us suppose that √2 is a rational number.

Then √2= a/b              ( a and b are coprime integers and b is not equal to 0.)

Squaring both sides

2= a^2/b^2

2b^2=a^2

2|a^2

So, 2|a        (i)               (if p divides a^2, then p also divides a.)

Let a=2c, where c is any integer

Squaring both sides

a^2=(2c)^2

or 2b^2=4c^2            (as 2b^2=a^2)

or b^2=2c^2

Thus, 2| b^2

and 2| b          (ii)    (if p divides a^2, then p divides a.)

From (i) and (ii),

We get that a and b have a common factor other than 1,i.e., 2.This contradicts our assumption that a and b are coprimes and the fact that $\sqrt{2}$ is rational.

Thus, √2 is irrational.

Now let us suppose that 5+3√2 is rational.

Then 5+3√2 = p/q       (p and q are coprime integers and q is not equal to 0.)

3√2 =p-5q/q

or √2 =p-5q/3q

We know that p,5,3 and q are integers. So, p-5q/3q must be a rational number. But we have also proved above that √2  is irrational. Rational cannot be equal to irrational. This contradicts our assumption that 5+3√2 is rational.

Thus, 5+3√2 is irrational.

Hence proved.