Question

prove that 3√2 is an irrational number.​

Answer 1

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Let us assume, to the contrary, that 3

2

is

rational. Then, there exist co-prime positive integers a and b such that

3

2

=

b

a

⇒

2

=

3b

a

⇒

2

is rational ...[∵3,a and b are integers∴

3b

a

is a rational number]

This contradicts the fact that

2

is irrational.

So, our assumption is not correct.

Hence, 3

2

is an irrational number.

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Answer 2

Answer:

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:

#Hope you have satisfied with this answer.