Question

prove
that these was irrational
√7​

Answer 1

let us assume that √7 be rational.

then it must in the form of p / q [q ≠ 0] [p and q are co-prime]

√7 = p / q

=> √7 x q = p

squaring on both sides

=> 7q2= p2 ------ (1)

p2 is divisible by 7

p is divisible by 7

p = 7c [c is a positive integer] [squaring on both sides ]

p2 = 49 c2 --------- (2)

Subsitute p2 in equ (1) we get

7q2 = 49 c2

q2 = 7c2

=> q is divisible by 7

thus q and p have a common factor 7.

there is a contradiction

as our assumsion p & q are co prime but it has a common factor.

So that √7 is an irrational.

.

.

$itzDopeGirl❣$

Answer 2

Answer:

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