Prove that for every $x,y\in\mathbb{R} $ such that $x,y>0$ and that fulfill $x\cdot y=1$, the following holds:

$$x+y=2 \quad iff \quad x=y=1$$

Steps I took:

$$\frac { xy }{ x } =\frac { 1 }{ x } \Rightarrow y=\frac { 1 }{ x } $$

$$1+\frac { 1 }{ 1 } =2$$

Is this good enough? It feels like something is missing.

Constructing my proof using input from answers below:

Assume: $x,y\in\mathbb{R}$ such that $x,y>0$ and $xy=1$

1) $x+y=2\Rightarrow x=y=1$

$y=2-x\Rightarrow x+2-x=2$; therefore, $2-x=1$

2) $x=y=1\Rightarrow x+y=2$


$x(2-x)=1\Rightarrow 2x-x^{ 2 }=1\Rightarrow x^2-2x+1=0$

$(x-1)^2=0\Rightarrow x=1$

$y=2-x\Rightarrow y=1$

  • 1
    $\begingroup$ You have shown that $x=1,y=1$ satisfies the equations. You have not shown that nothing else does. Using your approach, one can. Substituting, we get $x+\frac{1}{x}=2$. Show this equation forces $x=1$. It is somewhat more pleasant to subsitute $y=2-x$ in the equation $xy=1$, no fractions. $\endgroup$ – André Nicolas Oct 21 '15 at 16:29
  • $\begingroup$ @AndréNicolas How do I go about proving that nothing else can? $\endgroup$ – Cherry_Developer Oct 21 '15 at 16:31
  • $\begingroup$ How did the equation simplify to $(x-1)^2=0$ ? I feel like I am missing something here. $\endgroup$ – Cherry_Developer Oct 21 '15 at 16:37
  • $\begingroup$ Oh!. I forgot about the $xy=1$ in the initial claim. No wonder I got so confused $\endgroup$ – Cherry_Developer Oct 21 '15 at 16:40
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Cherry_Developer Oct 21 '15 at 18:24

The approaches described in the answers already posted are the most natural ones. But the following may be of interest. By expanding, we can show that for any $x$ and $y$ we have $$(x-y)^2+4xy=(x+y)^2.\tag{1}$$ If $xy=1$ and $x+y=2$, then substituting in (1) we get $(x-y)^2=0$, and therefore $x=y$. Now from $x+y=2$ we find that $x=y=1$.

So if our two given equations hold, then $x=y=1$. And it is easy to verify that $x=y=1$ does indeed satisfy the equations.

  • $\begingroup$ I will add the actual proof to my original post. Please let me know what to add/fix/delete. I just started proof writing about a week ago and it's proving to be very difficult. I can't seem to organize my thoughts in a way that proves these "trivial" things. $\endgroup$ – Cherry_Developer Oct 21 '15 at 17:10
  • $\begingroup$ Can you please look over the proof I have written up underneath my original question? $\endgroup$ – Cherry_Developer Oct 21 '15 at 19:49
  • $\begingroup$ What is (currently) written is, I am afraid, some distance from being satisfactory. You use too many symbols and too few words. Let's first look at part 2). You are trying to show that if $x=y=1$ then $x+y=2$. This is completely trivial, it says that $1+1=2$. The other direction, the one you call 1), is the one where we have to write a little, and where the stuff you wrote under 2) is relevant. (to be continued) $\endgroup$ – André Nicolas Oct 21 '15 at 20:04
  • $\begingroup$ (Cont) To show that $x+y=2$ implies, in the presence of $xy=1$, that $x=y=1$ do this. Suppose that $x+y=2$. Then $y=2-x$. Substituting for $y$ in the equation $xy=1$, we obtain $x(2-x)=1$. This simplifies to $x^2-2x+1$, or equivalently $(x-1)^2=0$. But $(x-1)^2=0$ if and only of $x-1=0$, or equivalently if and only if $x=1$. Finally, from $x=1$ and $x+y=2$, we conclude that $y=1$. $\endgroup$ – André Nicolas Oct 21 '15 at 20:09
  • 1
    $\begingroup$ It takes a while. But as a start, it is important to keep the logic straight. If you want to prove that $x=y=1\implies x+y=2$, then you can suppose that $x=y=1$ and then all you need is to verify that $x+y=2$. For the other direction, you want to prove that if $x+y=2$, then $x=y=1$ is forced. This is where you need an argument somewhat like the one you gave in 2). $\endgroup$ – André Nicolas Oct 21 '15 at 20:51

For the non-trivial way, suppose $x+y=2$, then $y=2-x$ and hence $x.(2-x)=1$, so $x^2-2x+1=0$. Now you can proceed from here.


In order to prove bi-conditional statements, iff, we must show that "both" directions are true. That is, if we have the statement "$A$ iff $B$", then we must show that "if $A$, then $B$. AND if $B$, then $A$", only then could we conclude $A$ iff $B$.

So, we first suppose that we have found $x,y \in \mathbb{R}$ such that $x,y > 0$ and $xy=1$. Then, we must show that (1) if $x +y =2 $, then $x=y=1$ AND (2) if $x=y=1$, then $x+y=2$.

Direction (2) is painfully obvious, but use the comments from seeker for (1). What you have done in your post is show that if $xy=1$, then one number must be the multiplicative inverse of the other. This is fine, and in fact, true. However, this does not help prove your statement.

  • $\begingroup$ I will add the actual proof to my original post. Please let me know what to add/fix/delete. Proof writing is proving to be very difficult. Even for such seemingly simple things. $\endgroup$ – Cherry_Developer Oct 21 '15 at 17:02

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