I want to prove that $x\sqrt{y}=y\sqrt{\dfrac{x^2}{y}}$; I've proven it to myself via calculator (brute forcing it) when $x,y>0$, and this is my proof: $$\begin{align*} x\sqrt{y} &= y\sqrt{\dfrac{x^2}{y}}&&\text{Conjecture}\\ &= y\dfrac{\sqrt{x^2}}{\sqrt{y}}&&\text{By Theorem 2.2}\\ &= y\dfrac{x}{\sqrt{y}}&&\because\sqrt{x^2}=x\\ &= \dfrac{xy}{\sqrt{y}}&&\text{Simplification}\\ x\sqrt{y}\cdot\dfrac{1}{x} &= \dfrac{xy}{\sqrt{y}}\cdot\dfrac{1}{x}&&\text{Multiplication}\\ \sqrt{y} &= \dfrac{y}{\sqrt{y}}&&\blacksquare \end{align*}$$


  1. In a proof, does it suffice to end with a truism as a result of the conjecture to prove the conjecture?
  2. Is my proof correct (the steps I took to get there)?
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    $\begingroup$ It's not true unless $x,y>0$. $\endgroup$ – Thomas Andrews Jan 3 '15 at 23:44
  • $\begingroup$ @ThomasAndrews Ah, yes; I forgot to include that. $\endgroup$ – Conor O'Brien Jan 3 '15 at 23:44
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    $\begingroup$ Not clear why the first step is "by definition." Anyway, you don't prove by starting with your conjecture and working backwards. That's how you might derive a proof, but the proof has to go forwards. So you'd start with $\sqrt{y}=\frac{y}{\sqrt{y}}$ and move in the other direction. $\endgroup$ – Thomas Andrews Jan 3 '15 at 23:45
  • $\begingroup$ I would say that if you are not sure if what you've done proves the conjecture, then it is not a proof, unless it happens to be a proof by pure accident. $\endgroup$ – Alexey Jan 3 '15 at 23:47
  • $\begingroup$ @ThomasAndrews By that, I simply am referencing the property of it, that, $\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$; on the second comment: Then how might I prove the original conjecture? $\endgroup$ – Conor O'Brien Jan 3 '15 at 23:47

You didn't write it properly. I believe you meant to write the following.

$$\color{blue}{x\sqrt{y} = y\sqrt{\dfrac{x^2}{y}}\tag{Conjecture}}$$

\begin{align*} x\sqrt{y} = y\sqrt{\dfrac{x^2}{y}}&\iff x\sqrt y= y\dfrac{\sqrt{x^2}}{\sqrt{y}}&&\text{By} \sqrt{\frac{x}{y}}\text{'s definition}\\ &\iff x\sqrt y= y\dfrac{x}{\sqrt{y}}&&\because\sqrt{x^2}=x\\ &\iff x\sqrt y= \dfrac{xy}{\sqrt{y}}&&\text{Simplification}\\ &\iff x\sqrt{y}\cdot\dfrac{1}{x} = \dfrac{xy}{\sqrt{y}}\cdot\dfrac{1}{x}&&\text{Multiplication}\\ &\iff \sqrt{y} = \dfrac{y}{\sqrt{y}}&&\blacksquare \end{align*}

The proof itself is correct, but, as it has been pointed out in the comments, the first justification is wrong. this answers the second question.

Answering the first question, yes, that is enough, you've proved that the conjecture is true if, and only if, something which is always true, is true, therefore the conjecture is always true.
You only need the direction going from bottom to top, though, so you can replace all the $\iff$'s by $\Longleftarrow$'s to get $$\sqrt y=\dfrac y{\sqrt y}\implies x\sqrt y=y\sqrt{\dfrac{x^2}{y}}.$$

Since the above statement is true (because of the given proof) and because the antecedent of the conditional statement is true, then the consequent can't be false.

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  • $\begingroup$ Does that mean I should simply reverse the steps I took, fixing the mistakes I made? $\endgroup$ – Conor O'Brien Jan 4 '15 at 0:01
  • $\begingroup$ Assuming you actually wanted to write what I typed above (please confirm this), then you just need to replace the $\iff$'s with $\Longleftarrow$'s, you don't even need to write in the reverse order. Inverting the order will make for a more natural reading, but it's not necessary. Edit: Mathematics must be parsed all at once. That you read from left to write and top to bottom is a cultural condition. Both $$\begin{align} A&\Longleftarrow B\\ &\Longleftarrow C\end{align}$$ and $$\begin{align} C&\implies B\\ &\implies C\end{align}$$ state the same. $\endgroup$ – Git Gud Jan 4 '15 at 0:03
  • $\begingroup$ Yes, that works (that is, what typed is what I want). Thanks! $\endgroup$ – Conor O'Brien Jan 4 '15 at 0:08
  • $\begingroup$ @ConorO'Brien You're welcome. Please note the edit to the above comment. $\endgroup$ – Git Gud Jan 4 '15 at 0:09
  • $\begingroup$ I looked at it, and it slightly confuses me; Why is the second example equal to the first? In the second, there is no $A$, but there is in the first. Is that a typo, or is it meant? $\endgroup$ – Conor O'Brien Jan 4 '15 at 0:10

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