Question

prove that √7 is a irrational number​

Answer 1

$\huge{ \underline{ \underline {\mathfrak {\red{ Answer : }}}}}$

let us assume that √7 be rational. then it must in the form of p / q.

• As definition of rational number says.. P is whole number q is non zero whole number.. And p and q is simplest ratio which is expressed.. That means there exists no prime factor common in p and q.

√7 = p / q

√7 x q = p

squaring on both sides

7q² = p² ------1.

p is divisible by 7

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

p²= 49c²

subsitute p² in eqn(1) we get

7q² = 49 c²

q² = 7c²

q is divisble by 7

thus q and p have a common factor 7.

there is a contradiction to our assumption

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

so that √7 is an irrational.

Happy Learning..!!

Answer 2

$\huge\frak \red{\fcolorbox{red}{lavender} { \ \: \: Here\ is\ your\ answer\: \: }}$

Let assume that √7 is rational.

As we know that rational number can be in form a/b. ( here a and b are co prime means they have only 1 as common factor).

√7 = a/b

√7 = a/b

• On squaring both sides

7 = a²/b²

7b² = a²

b² = a²/7

• Here 7 divides a² so for it will also divide a .

a²/7 = a/7

Let a = 7c

• On squaring both side

a² = 49c²

• We know that a² = 7b²

7b² = 49c²

b² = 7c²

b²/7 = c²

• Here 7 divides b² so for 7 divides b.

Now, we find that 7 is common which divide both a and b but this is contradiction because a and b are co prime they don't have common factor other than 1. So the assumption is wrong.

Hence √7 is irrational.

This answer is not copied. I am the one who answered it first. You can see it here ↓

https://brainly.in/question/36984005