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 = m2 and b = n2. Then
a*b = m2*n2 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 = m2 and b = n2, 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.