Question

prove that 3+7root2 is irrational​

Answer 1

Answer:

sum of two irrational no. is always irrational. Take an example of underroot 2 and underroot 2 there sum comes out to be 2 underoot 2 Assume thus no. To be rational no. shift 2 on RHS underroot 2 is irrational no.but p/ 2q is irrational no. and write the solution which is written in pic.

Or, you can do it by another method given below:

Let us assume 3+7 √2 as rational number

3+7√2 =a\b

√2 = a/7b-3

Here,

a/7b-3 is rational number so

3+7√2 is also rational number

But √2 is irrational

This contradiction had arrisen due to our incorrect assumption

This contradicts the fact that 3+7√2 is irrational.

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Answer 2

Answer:

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Step-by-step explanation:

✏Let assume 3+7√2 is a rational number

3+7√2=a\b

√2=a\7b-3

Here,

a\7b -3 is rational number so

3+7√2 is also a rational number

but √2 Is rational

So there is a contradiction had arisen due to our incorrect assumption

This contradiction fact that 3+7√2 is irrational

Hope it's helpful↑

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