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I should do it by contradiction. Would finding a counterexample work? For example:

Suppose $\sqrt{x}+\sqrt{y}=\sqrt{x+y}$. Then $\sqrt{4}+\sqrt{9}=\sqrt{4+9}$, which means $2+3=\sqrt{13}$ which is a contradiction.

Is this a correct (even though no beautiful) proof?

pipey
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4 Answers4

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If $\sqrt x+\sqrt y=\sqrt{x+y}$,

then $x+2\sqrt{x}\sqrt{y}+y=x+y\Longrightarrow \sqrt x\sqrt y =0$

which is impossible since $x,y$ are positive reals. Contradiction.

  • Again, as i said, i did it like this as well, but i just wanted to know if in this other way it would have counted anyway – pipey Sep 30 '20 at 09:47
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    I think this should do since there don’t seem to exist any problem in it. In addition, it is just a normal proof by contradiction. –  Sep 30 '20 at 09:49
  • @pipey, one counter-example indeed is sufficient and I think it is fine, but the contradiction and counter-example don't mean the same thing. In a proof by contradiction, you don't explicitly give an example the statement doesn't hold for. – PinkyWay Sep 30 '20 at 09:49
  • @Invisible so the contradiction i gave in the proof above it's not a "proper contradiction"? – pipey Sep 30 '20 at 09:51
  • @pipey: To disprove a statement, a single counter example is sufficient. To prove a statement, you need to prove it for all cases. If you state, 'All swans are white' you need to produce only one black swan to disprove that statement. If you want to prove that statement 'All swans are white', you need to produce all swans in the universe and demonstrate that they are white. – vvg Sep 30 '20 at 10:31
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Assume it is true. Square both sides. You'll get $x+y+2\sqrt{xy}= x+y$.

$\implies \sqrt{xy} = 0$ $\implies xy=0$

$ \implies x=y=0$

or $ x=0, y \neq 0$

or $ y=0, x \neq 0$

So this statement isn't true for all $x,y$, but only the above 3 cases.

Syed Bukhari
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Whether a single example counts as a complete, correct proof depends on the tacit quantifier behind the assertion "$\sqrt x+\sqrt y\not=\sqrt{x+y}$ where $x$ and $y$ are positive real numbers." If the careful formulation is

$$\exists x,y\in\mathbb{R}^+: \sqrt x+\sqrt y\not=\sqrt{x+y}$$

then yes, a single example suffices. But if the careful formulation is

$$\forall x,y\in\mathbb{R}^+: \sqrt x+\sqrt y\not=\sqrt{x+y}$$

then a single example is not enough (and a proof along the lines of the other answers here is required).

To move the negation around, the first formulation says that the equation $\sqrt x+\sqrt y=\sqrt{x+y}$ (with $x,y\in\mathbb{R}^+$) is not an identity, whereas the second formulation says the equation has no solutions.

Barry Cipra
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Overkill?

Let $x,y \ge 0$, real.

Then set $a:=√x;$ $b:=√y;$

We have

LHS: $a+b;$

RHS: $\sqrt{a^2+b^2}=\sqrt{(a+b)^2-2ab}\le $

$\sqrt{(a+b)^2}=a+b.$

Equality: $\iff$ $ab=0,$ i.e. $a=0$ or $b=0.$

Used: $f(x)=√x$ is an increasing function.

Peter Szilas
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