Question

prove that 3√2 is irrational​

Answer 1

Answer:3+√2 = a/b ,where a and b are integers and b is not equal to zero .. therefore, √2 = (3b - a)/b is rational as a, b and 3 are integers.. ... So, it concludes that 3+√2 is irrational..

prove :

Let 3+√2 is an rational number.. such that

3+√2 = a/b ,where a and b are integers and b is not equal to zero ..

therefore,

3 + √2 = a/b

√2 = a/b -3

√2 = (3b-a) /b

therefore, √2 = (3b - a)/b is rational as a, b and 3 are integers..

It means that √2 is rational....

But this contradicts the fact that √2 is irrational..

So, it concludes that 3+√2 is irrational..

hence proved..

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Let us assume that $3\sqrt{2}$ is a rational number.

therefore, it can be shown in the form of p / q.

Hence, $3\sqrt{2}$=p/q

            ⇒p/3q

Here, root 2 is irrational number.

         p/3q is a rational number.

As, rational ≠ irrational

So, our assumption is wrong.

Hence proved, 3√2 is an irrational number.

 

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