Question

prove that √7 is an irrational​

Answer 1

$\underline\bold{\huge{SOLUTIONS:\: :}}$

Let us assume that √7 can be expressed in the form p/q where q is not equal to

zero.

And p and q has no common factor other than 1.

So,√7=p/q

Square both sides we got

7=p^2/q^2

7q^2=p^2...(1)

So, 7 divide p

reasons:if a no. say X can divide no.say

say y^2 then X can divide y

so, let 7m=p

squaring both sides we got

49m^2=p^2

replacing the p^2 in equation 1 we got

7q^2=49m^2

q^2=7m^2

so, 7 divide q

Now since 7 is the factor of both p and q

Hence this contradict the fact that √7 is a rational no.

$\underline\bold{\huge{So, √7 is an irrational number\: :}}$

$\red{\huge{Hi \:follow\: me\: guys}}$

Answer 2

Answer:

Step-by-step explanation:

Lets assume that √7 is rational number. ie √7=p/q.

suppose p/q have common factor then

we divide by the common factor to get √7 = a/b were a and b are co-prime number.

that is a and b have no common factor.

√7 =a/b co- prime number

√7= a/b

a=√7b

squaring

a²=7b² .......1

a² is divisible by 7

a=7c

substituting values in 1

(7c)²=7b²

49c²=7b²

7c²=b²

b²=7c²

b² is divisible by 7

that is a and b have atleast one common factor 7. This is contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.

√7 is irrational✔✔

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