Question

Prove that √7 is an irrational number.​

Answer 1

Step-by-step explanation:

Let us assume that

✓7

is rational. Then, there exist co-prime positive integers a and b such that

√7

=

b

a

⟹a=b

√7

Squaring on both sides, we get

a

2

=7b

2

Therefore, a

2

is divisible by 7 and hence, a is also divisible by7

so, we can write a=7p, for some integer p.

Substituting for a, we get 49p

2

=7b

2

⟹b

2

=7p

2

.

This means, b

2

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

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

But, this contradicts the fact that a and b are co-prime.

Thus, our supposition is wrong.

Hence,

√7

is irrational.

Answer 2

Answer:

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