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$f$ and $g$ are functions of real variables, strictly increasing and strictly decreasing respectively on $\Bbb R$ (both surjections), I'm asked to prove that there exists at most one solution to the equation $f(x)=g(x)$.

My attempt: Suppose $f(x)=g(x)=m,$ for some $x\in \Bbb R \implies \exists a,b \in \Bbb R | a\lt x\lt b$. Then : $$f(a)\lt f(x)\lt f(b)$$ and $$-g(a)\lt -g(x)\lt -g(b).$$ Hence $$f(a)-g(a)\lt f(x)-g(x)\lt f(b)-g(b).$$ So if we can prove that $f(a)-g(a)\lt 0$ and $f(b)-g(b)\gt 0$. Hence for $f(a)-g(a)=p\lt 0$ and $f(b)-g(b)=q\gt 0$, we will have $p\lt 0\lt q \equiv \top$, but I couldn't prove it. Any help would be appreciated.

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It seems you have already proven the part you intended in $$f(a)-g(a)\lt f(x)-g(x)\lt f(b)-g(b)$$ just notice that $f(x)-g(x)=0$ (why?). So this way you have $f(a)<g(a)$ for $a<x$ and $f(b)>g(b)$ for $b>x$, which completes the proof (equality cannot occur).

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  • $\begingroup$ Will that complete the proof ? @Sil $\endgroup$ – FuzzyPixelz Dec 30 '16 at 22:48
  • $\begingroup$ @Mahmoud Well it only proves the part you intended, i.e. that $f(a)-g(a)<0<f(b)-g(b)$, I am not sure where you are going with that afterwards. I have updated my answer to give some way to consider. $\endgroup$ – Sil Dec 30 '16 at 22:55
  • $\begingroup$ I couldn't separate the cases. @Sil $\endgroup$ – FuzzyPixelz Dec 30 '16 at 22:57
  • $\begingroup$ I just realized that this way you have shown that for $b>x$ you have $f(b)>g(b)$ and for $a<x$ that $f(a)<g(a)$ so yes, this completes the proof (there will never be equality). $\endgroup$ – Sil Dec 30 '16 at 23:04
  • $\begingroup$ @Mahmoud You are welcome! Just one more comment, you showed this for all $a,b$ (the $\exists a,b \in \Bbb R$ is incorrect in your statement). $\endgroup$ – Sil Dec 30 '16 at 23:07
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Hint: If there were two different points, either $f$ increasing or $g$ decreasing would be wrong.

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