Answer:

let us assume that √7 be rational. thus q and p have a common factor 7. as our assumsion p & q are co prime but it has a common factor. So that √7 is an irrational

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Verified answer

[tex]\colorbox{lightpink}{★ANSWER ★}[/tex]

[tex]suppose \: that \sqrt{7} \: is \: rational[/tex]

[tex]then, \sqrt{7} = \frac{a}{b} ;where \: a \: and \: b \: are \: co - prime \: integers[/tex]

Squaring on both the side,we get

[tex]7 = \ \frac{ {a}^{2} }{{b}^{2} } \\ a {}^{2} = 7b {}^{2} [/tex]

Hence, 7 is a factor of a^2.

since 7 is a prime numbers, by theorem 1.3,7 is also factor of a

let a=7c,where c is an integer.

[tex]a {}^{2} = 49 {c}^{2} \\ 7 {b}^{2} = 49 {c}^{2} \\ {b}^{2} = 7 {c}^{2} [/tex]

Hence,7 is a factor of bd^2

since 7 is a prime numbers, by theorem 1.3,7 is also a factor of b

thus, a and b have common factor 7, which contradicts our assumption

that a and b are co-prime integer

Hence, our assumption is incorrect

SO WE CONCLUDE THAT THE sqrt{7} IS

IRRATIONAL

[tex]\sf \colorbox{yellow} {ANSWER BY ACHALMUCHHAL2}[/tex]