Question

prove that 12√13 is irrational (real numbers)​

Answer 1

Answer:

this is solution by given answer.

Step-by-step explanation:

Answer:

root 2+ root 7

root 2+root 7=a (where a is an integer)

squaring both sides

(root 2+root 7)^2=(a)^2

(root 2)^2+(root 7)^+2(root 2)(root 7)=a^2

2+7+2 root 14=a^2

9+2 root 14 =a^2

2 root 14=a^2-9

root 14=a^2-9/2

since a is an integer therefore a^2-9/2 is also an integer and therefore root 14 is also an integer but integers are not rational numbers therefore root 2+root 7 is an irrational number.

proved.

Answer 2

$\huge\mathfrak{{\fcolorbox{gold}{gold}{Answer!}}}$

Let us assume that $\large\bold{12{\sqrt{13}}}$ is rational...

Then we get, $\large\bold{12{\sqrt{13}}={\frac{p}{q}}}$ , where p and q are integers, and q≠0

$\large\implies\bold{{\sqrt{13}}={\frac{p}{q}}×{\frac{1}{12}}={\frac{p}{12q}}}$ represents a rational number

Where $\large\bold{{\sqrt{13}}}$ is irrational. Thus LHS ≠ RHS

$\large\implies$ our assumption is wrong, $\large\bold{12{\sqrt{13}}}$ is not rational it is Irrational...

$\huge\bold\orange{Brainliest~Please!}$