Question

prove that 6+3√2 is irrational​

Answer 1

6 + 3√2 is irrational number

__________ [PROVE]

Solution:

• Let us assume that 6 + 3√2 is irrational number.

=> 6 + 3√2 = $\dfrac{a}{b}$

Here, a and b are co-prime numbers.

=> 3√2 = $\dfrac{a}{b}$ - 6

=> 3√2 = $\dfrac{a\:-\:6b}{b}$

=> √2 = $\dfrac{a\:-\:6b}{3b}$

Here;

$\dfrac{a\:-\:6b}{3b}$ is rational number.

So, √2 is also a rational number.

But we know that √2 is irrational number.

So, our assumption is wrong.

6 + 3√2 is irrational number.

Hence, proved.

______________________________

Answer 2

Let assume 6+3√2 be a rational number

6+3√2=a/b___________where a and b are co prime and their HCF is 1

√2=a/b-6/3

√2=3a-6b/3b

√2=3a-2b

Where 3a-2b is rational number therefore √2 also be a rational number

But we know that√2 is irrational number

Hence 6+3√2 is irrational number

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