[tex]\huge \blue{ \bold { \pmb{ \fcolorbox{gray}{black} { {\ \ddag \: \ \red{QUESTION}\ \ddag \: }}} }}[/tex]

July 7, 2021

By Iris

2 thoughts on “[tex]\huge \blue{ \bold { \pmb{ \fcolorbox{gray}{black} { {\ \ddag \: \ \red{QUESTION}\ \ddag \: }}} }}[/tex]<br /><br /><b”

1. Explanation:

   as our assumsion p & q are co prime but it has a common factor. So that √7 is an irrational.

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2. 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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