Re: is this proof correct?

Quote:

Originally Posted by

**ga04** Theorem: for all non-negative real numbers a and b, SQRT(a*b) = SQRT(a)*SQRT(b)

Proof: suppose a and b are any non negative real numbers. Then there exists unique non negative real numbers m and n such that a = m^{2} and b = n^{2}. Then

a*b = m^{2}*n^{2} by substitution

= (m*n)^{2} by laws of exponents.

Then by taking the square root of both sides

SQRT(a*b) = m*n

but because a = m^{2} and b = n^{2}, it follows that m = SQRT(a) and n = SQRT(b), and so , by substitution

SQRT(a*b) = SQRT(a)*SQRT(b)

Q.E.D

Thanks in advanced for your comments :)

I think this is an absolutely correct proof because the problem is just to show the result holding for "Non-Negative" reals only...

Re: is this proof correct?

Yes of-course this proof is correct, as this proof is satisfying the given theorem.