# is this proof correct?

• May 18th 2012, 10:42 AM
ga04
is this proof correct?
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

• May 21st 2012, 07:49 PM
Sarasij
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