Question

prove that 3 + √3 is irrational​

Answer 1

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}$

GIVEN:

$A\:number\:3+\sqrt{3}.$

TO PROVE :

$It \:is \:irrational.$

PROOF:

On the contrary let us assume that it is a rational number,

So, it can be expressed in the form of$\dfrac{p}{q}$where p and q are integers and q≠0.Also, p and q are co-primes.

______________________________________

ATQ,

=>$\dfrac{p}{q}=3+\sqrt{3}$

On squaring both sides, we have,

=>$(\dfrac{p}{q}) ^{2}=(3+\sqrt{3}) ^{2}$

=>$\dfrac{p^{2}}{q^{2}}=(3) ^{2}+(\sqrt{3})^{2}+2×3×\sqrt{3}$

$\large\green{\boxed{(a+b) ^{2}=a^{2}+b^{2}+2ab}}$

=>$\dfrac{p^{2}}{q^{2}}=9+3+6\sqrt{3}$

=>$\dfrac{p^{2}}{q^{2}}=12+6\sqrt{3}$

=>$\dfrac{p^{2}}{q^{2}}-12=6\sqrt{3}$

=>$\dfrac{p^{2}-12q^{2}}{q^{2}}=6\sqrt{3}$

=>$\dfrac{p^{2}-12q^{2}}{6q^{2}}=\sqrt{3}$

But, now,we have arrived at a contradiction,LHS is a rational number but RHS is a rational number.

Therefore our assumption was wrong $3+\sqrt{3}$ is a irrational number.

$\huge\orange{\boxed{HENCE\:\:PROVED}}$

Answer 2

Answer:

Step-by-step explanation:

let us assume $3+\sqrt{3$ as a rational number

now let us take a and b where a and b are two integers and b$\neq$0

now let us a and b as the form of a/b

=) $3+\sqrt{3}$=a/b

=) $\sqrt{3}$=a/b-3

=) $\sqrt{3}$=a-3b/b

so the a-3b is a rational number and then $\sqrt{3}$ will also be a rational number

(as per assumption)

however $\sqrt{3\\}$ is a irrational number

therefore we arrives at a contradiction due to our incorrect assumption

so we conclude $3+\sqrt{3}$ as an irrational number