Question

given that root 2 is irrational prove that(5+3 root 2) is an irrational number

Answer 1

let 5+3√2 is rational no.
so , their exit co.prime intezer p and q , q is not equal to 0
then ,
5+3√2 = p/q
3√2 = p/q - 5
3√2= p - 5q divided by q
the contradaction for fact 5+3√2 is rational
our assumption is wrong
5+3√2 is irrational no.
proved

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Answer 2

Answer:

Given : $\sqrt{2}$ is irrational number.

To prove : $(5+3\sqrt{2})$ is an irrational number.

Assumption:

Let us assume $(5+3\sqrt{2})$ is a rational number.

Proof:

As $(5+3\sqrt{2})$ is rational.

They must be in the form of p/q where $q\neq 0$, and  p and q are co prime.

Then,

$(5+3\sqrt{2})=\frac{p}{q}$

$3\sqrt{2}=\frac{p}{q}-5$

$3\sqrt{2}=\frac{p-5q}{q}$

$\sqrt{2}=\frac{p-5q}{3q}$

We know,

$\sqrt{2}$ is irrational number

$\frac{p-5q}{3q}$ is a rational number.

And $\text{Rational}\neq \text{Irrational}$

Therefore, our assumption is wrong.

$(5+3\sqrt{2})$ is an irrational number.