Prove that 3√2 is irrational.
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Answered by
3
let 3√2 is rational number
therefore it is in the form p/q where p and q are co prime numbers
3√2=p/q
√2 = p/3q
therefore l.h.s. is irrational no. and rhs is rational no.
but it is a contradiction
therefore our supposition is wrong
therefore it is a irrational number
therefore it is in the form p/q where p and q are co prime numbers
3√2=p/q
√2 = p/3q
therefore l.h.s. is irrational no. and rhs is rational no.
but it is a contradiction
therefore our supposition is wrong
therefore it is a irrational number
Answered by
1
let us take contradiction that 3√2 is rational number.
so if 3√2 is a rational number then
let 3√2=a/b,where a and b are co primes
then
3√2=a/b
√2=a/3b
here in the R.H.S side
that is a/3b is a rational number but in the L.H.S side √2 is irratioanal
so 3√2 is a irrational number..................hope it helped u:))))))
so if 3√2 is a rational number then
let 3√2=a/b,where a and b are co primes
then
3√2=a/b
√2=a/3b
here in the R.H.S side
that is a/3b is a rational number but in the L.H.S side √2 is irratioanal
so 3√2 is a irrational number..................hope it helped u:))))))
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