Question

prove that √7 is a irrational​

Answer 1

Step-by-step explanation:

hey mate we can do this by taking an assumption that it is rational .

it's could be like this:-

Let us assume that

$step - 1 = >$

$\sqrt{7}$

is rational.

$step - 2 = >$

Then, there exist co-prime positive integers a and b such that =>

$\sqrt{7 } = \frac{a}{b}$

$a = b \sqrt{7}$

$step - 3 = >$

by squaring on both sides we get :-

$= > {a}^{2} = {b}^{2} 7$

Therefore,

${a}^{2}$

is divisible by 7 and hence, a is also divisible by 7

$step - 4 = >$

so we can also write it as :-

a=7p. (p is any integer)

$step - 5 = >$

substitute for a

we get,

$49 {p}^{2} = 7 {b}^{2}$

${b}^{2} = 7 {p}^{2}$

$step - 6 = >$

This means,

${b}^{2}$

is also divisible by 7 and so, b is also divisible by 7.

$step - 7 = >$

Therefore, a and b have at least one common factor, i.e., 7.

but, this contradict fact that this is not possible.

$final \: step = >$

by which our assumption become wrong

and it's proves that

$\sqrt{7}$

is irrational number.

Answer 2

Answer:

$let \sqrt{7 \: } be \: rational \: number \\ \\ \therefore \: \sqrt{7} = \frac{a}{b \:} \: ( \: where \: a \: and \: b \: are \: co - prime \: and \: b \: ≠0) \\ \\ \:\implies7 = \frac{ {a}^{2} }{ {b}^{2} } \\ \\ \implies7 {b}^{2} = {a}^{2} \\ \\ \therefore \: {a}^{2} \: is \: divisible \: by \: 7. \\ \\ hence \: a \: \: is \: also \: divisible \: by \: \: 7. \\ \\ again \: \: \: let \: a = 7c \\ \\ \implies {a}^{2} = {(7c)}^{2} \: \\ \\ \implies {7b}^{2} = {49c}^{2} ( \: putting \: {a}^{2} = {7b}^{2} ) \\ \\ \implies \: {b}^{2} = \frac{ {49c}^{2} }{7} \\ \\ \implies {b}^{2} = {7c}^{2} \\ \\ \therefore \: {b \: }^{2} is \: \: also \: divisible \: by \:7. \\ \\ hence \: \: b \: is \: also \: divisible \: by \: 7. \\ \\ but \: this \: contradicts \: the \: fact \: that \: a \: and \: b \: have \: no \: common \: factor \: other \: than \: 1. \\ as \: a \: and \: b \: have \: have \: 7 \: as \: a \: common \: factor. \\ hence \: \sqrt{7} \: \: is \: not \: rational. \\ \therefore \: \sqrt{7} \: is \: irrational.$