$P \implies Q$ using proof by contrapositive

Question

Let $x,y∈(0,1)$. Prove that if $x \ne y$, then $ \frac{x}{x^2+1} ≠ \frac{y}{y^2+1}$

Contrapositive: $\frac{x}{x^2+1} =\frac{y}{y^2+1} => x=y$

Suppose $¬Q$ happens

$\frac{x}{x^2+1} =\frac{y}{y^2+1}$

$x(y^2+1) = y(x^2+1)$

$xy^2+x = yx^2+y$

$x=y$

How do i conclude my solution?

$x=y$ or $xy=1$ But since $x,y\in(0,1),$ we do not have the second case. — Bumblebee, Jul 07 '17 at 22:47
That's it as far as I can see. Since you've shown $\frac{x}{x^2+1} =\frac{y}{y^2+1} => x=y$, you can conclude $\frac{x}{x^2+1} \neq\frac{y}{y^2+1} <= x\neq y$. — Shuri2060, Jul 07 '17 at 22:47
Using the contrapositive? $(P\implies Q) \iff (\neg Q \implies \neg P)$. That's literally what it says. — Shuri2060, Jul 07 '17 at 22:49
$P \implies Q$ and $\neg Q \implies \neg P$ are tautological. Proving one of them proves both. — dromastyx, Jul 07 '17 at 22:53
@dromastyx do i need to worry about the x,y∈(0,1) in my answer? — TheGamer, Jul 07 '17 at 22:54
@TheGamer Yes, in the last step of your proof, as shown by Siong Thye Goh — dromastyx, Jul 07 '17 at 22:56
I'd also suggest putting $\implies$ between each line to make it clear how the proof is to be followed through. — Shuri2060, Jul 07 '17 at 22:59
@dromastyx can i update my answer so ik if thats how u want me to do it — TheGamer, Jul 07 '17 at 23:01
@TheGamer Apart from minor things like implications between lines that looks fine to me. — dromastyx, Jul 07 '17 at 23:07
Also, include the final line which is 'By the contrapositive $x\neq y \implies \frac{x}{x^2+1} ≠ \frac{y}{y^2+1}$', or something similar. — Shuri2060, Jul 07 '17 at 23:09

Siong Thye Goh · Accepted Answer · 2017-07-07T23:03:38.553

2

$$xy^2-yx^2 = y-x$$

$$xy(y-x)=y-x$$

$$(y-x)(xy-1)=0$$

Suppose $y -x \neq 0$, then we have $xy-1=0$, which means $xy=1$. This is impossible since $x,y \in (0,1)$. Hence $y-x=0$, that is $y=x$.

edited Jul 07 '17 at 23:03

answered Jul 07 '17 at 22:47

Siong Thye Goh

Small suggestion - factorise for complete clarity? $(xy-1)(y-x)=0$ – Shuri2060 Jul 07 '17 at 23:01
ah, sure, that certainly makes things clearer. Thanks. – Siong Thye Goh Jul 07 '17 at 23:02
@SiongThyeGoh where did u get this statement from (xy−1)(y−x)=0 – TheGamer Jul 07 '17 at 23:10
$xy(y-x)=(y-x)$ Hence, we have $xy(y-x)-(y-x) = 0$, then I just factorize $y-x$. – Siong Thye Goh Jul 07 '17 at 23:12
$ab=0$ means $a$ or $b$ must be $0$. ie. $(y−x)(xy−1)=0 \implies (y-x=0) \lor (xy-1=0)$ (Note that $\lor$ is the logical 'or' symbol) – Shuri2060 Jul 07 '17 at 23:13

score 0 · Answer 2 · answered Jul 07 '17 at 23:33

Let $f (x)=\frac {x}{x^2+1} $ for $0 <x <1$.

suppose there exist $(x ,y)\in (0,1)^2$ such that $f (x)=f (y) $ and $x\ne y $.

$f $ is differentiable at $[0,1] $, then by Rolle's Theorem,

$$\exists c\in (x,y)\; :\; f'(c)=0$$ but $$f'(c)=\frac {1-c^2}{(c^2+1)^2}\ne 0$$ since $0 <c <1$

hence $x=y $.

2 Answers2