Question

Mathematically prove followings:-

1. $\bold\green{\dfrac{0}{0} }$ = $\huge{\boxed{\bold{2}}}$

2. $\bold\green{\sqrt{2} \: is\:irrational}$

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Answer 1

Answer:

Q= prove that root 5 is an irrational no

Step-by-step explanation:

(ii ) solution

let us assume that

$\sqrt{5}$

is a rational no

now we find integers a and b

$such \: that \: \sqrt{5} = \frac{a}{b}$

because rational no is always in the form of p upon q

where a and b are two co prime no and b#0 means denominator never zero

now squaring both sides

$\sqrt{5} = \frac{a}{b}$

so

$( \sqrt{5} ) {?}^{2} =( \frac{a}{b} ) {?}^{2}$

now

${b}^{2} 5 = {a}^{2}$

${b}^{2} = a \: square \: upon \: 5$

a square is divisible by 5

so

a is also divisible by 5

let a is a = 5 m fire some integar m

$\sqrt{5} = \frac{5m}{?b}$

squaring both sides

$( \sqrt{5} {?}^{2} ) = (5m \: upon \: b \: \: square$

$= 5 = \frac{25m {?}^{2} }{b {?}^{2} }$

${b}^{2} 5 = 25 {m}^{2}$

${b}^{2} = \frac{25m {?}^{2} }{5}$

${b}^{2} = 5 {m}^{2}$

so b square is divisible by 5

hence

b is also divisible by 5

so

we assume that root 5 is a rational number but it is not a rational number because it is not right in the form of p upon q so if it not a rational number so it is a rational number

$\sqrt{5} is \: an \: irrational \: no$

sorry I not see your question pls put 2 at the place of 5

Answer 2

$\rm\underline\bold{Answer \purple{\huge{\checkmark}}}$

5

isanirrationalno