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We took 2 integers i.e, r& s

Therefore, √3= r/s ,[ p/q form]

√3= r/x/s/x [xis a common factor] , =a/b

squaring on both sides

√3 b2 = a2

3b2 a2

b2=a2/3

=3divides a

Similarly , we have to proove 3dvides b also

From above situation

It i clear that a& b are integers & √3 is rational

But bythe fat √3 Is irrtional

So our assumption is wrong.

Then, √3 = a/b

when we square both sides,

3 = a²/b², which implies,

3b² = a²

Since 3 is a factor of 3b², a² is divisible by 3. Now, if a² is divisible by 3 then 'a' is also divisible by 3.

Now, since 'a' is divisible by 3, 3 is a factor of 'a'. That is, a = 3k, where 'k' is some integer. Now, 3b² = a² ⇒ 3b² = (3k)² ⇒ 3b² = 9k². Dividing both sides by 3, we have b² = 3k² which means that b² is divisible by 3. This implies that 'b' is also divisible by 3.