For Mathematicians Who can make and solve proofs in Algebra!?

Question

I need to prove or disprove
square root of "a" plus the square root of "b"
is equal to
square root of "a" plus "b"

Its not that easy because if you substitute in "a" is equal to 16 and "b" is equal to 16 the problem doesnt work out I know that the the equality can be proven but I'm just not sure what other steps I should take to prove the equation true. Thanks For your help... Oh.. By the way.. I'm no really worried about the answer just about the steps and how this can be proven so if you dont know how to do it and are just gonna type in the answer just dont answer, again this is for true mathematicians who can actually prove things in algebra.

Answer 1

The proposition is

SQRT(a) + SQRT(b) = SQRT(a+b)

If the equality is true, then we can rais both sides to the second power, yielding

(SQRT(a) + SQRT(b))^2 = (a+b)

If we put c=SQRT(a) and d=SQRT(b) to simplify the equation, we have

(c + d)^2 = a+b
and expanding
c^2 + 2cd + d^2 = a+b

But c^2 = SQRT(a) ^2, and d^2 = SQRT(b) ^2, so in the end
c^2 = a and d^2 = b
so back substituting we have
a +2cd +b = a+b
and 2cd must be equal to zero, and that is possible only if either c or d is zero, and that is possible only if either a or b is zero.

Answer 2

Hi Math Guy. I'm a mathematician and prove things in algebra all the time. My job is to consider propositions and prove them if they are correct and disprove them if they are false. The rules of argument are set up in such a way that we can never prove and disprove the same statement. In fact if there ever was *one* single statement in the theory both proven and disproven, the rules of logic would say that *every* statement in mathematics is both proven and disproven, and the whole theory would be lost.

Let's take your proposition for example. If anyone comes up with a proof of it, no matter how complex or mind-bending, I could go through the proof and plug in 16 for "a" and 16 for "b" everywhere and eventually find a flaw in the argument.
Maybe the proof would involve dividing by 0 or inconsistent rules of taking square roots; who knows. But there is no way it would work, unless we add the hypothesis that "a" and/or "b" is 0.

Answer 3

(sqrt)a + (sqrt)b = (sqrt)[a + b]
Prove or disprove
Put a = 16, b = 4
(sqrt)16 + (sqrt)4 = 4 + 2 = 6
But (sqrt)[16 + 4] = (sqrt)20 which is irrational
Hence, this statement is wrong and has been disproved.

Answer 4

square both of those expressions.
you get: a + b + 2sqrt(a)sqrt(b) = a + b
which is not necessarily true, so the statement is false