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I would like to prove that a strictly decreasing function from $f:\Bbb R \to \Bbb R$ is one-to-one.

We want to show that show that $f(a) = f(b)$ implies $ a = b$ for all $a, b \in \Bbb R$.

One proof I saw online was as follows (although I did the same proof using contrapositive technique), but I just want to get better understanding as to why he did the proof as follows:

Proof:

Since the function is strictly decreasing, it means that if $ x \lt y \implies f(x) \gt f(y)$. To proof that it's one-to-one function, we need to prove that if $f(a)=f(b) \implies a=b$.

Let $f(a) = f(b)$.

Case 1: Consider when $a \lt b$, then this implies that $f(a) \gt f(b)$ since $f(x)$ is strictly decreasing. This implies that $f(a) \ne f(b) \therefore a\ge b $.

Case 2: Consider when $a \gt b$, then this implies that $f(a) \lt f(b)$ since $f(x)$ is strictly decreasing. This implies that $f(a) \ne f(b) \therefore a = b $.

Questions:

  1. It seems the proof that was used in the question is proof by cases, was not it?
  2. Why it was assumed, in Case 1, that $f(a) = f(b)$ although what is given in the question is that $f(x)$ is strictly decreasing?
  3. Why it was concluded ,in Case 1, that since $f(a) \ne f(b) \therefore a\ge b $?
  4. I assume that it was finally concluded that $\therefore a = b $ is because no other scenarios left as to why $f(a) = f(b)$ except by equality of $a$ and $b$.
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    $\begingroup$ As you note, to prove $f$ is injective, it suffices to show that $f(a)=f(b)\implies a=b$. The proof assumes $f(a)=f(b)$, then deduces that $a<b$ and $a>b$ are both impossible (so yes, I guess technically the proof splits into two cases). Hence we must have $a=b$, proving what we want. $\endgroup$
    – jlammy
    Jan 15, 2021 at 16:29
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    $\begingroup$ The definition of one-to-one is that if $f(a)=f(b)$, then we must have $a=b$. So the obvious way to prove $f$ is one-to-one is to suppose that $f(a)=f(b)$ is true, then deduce that $a=b$ must be true. $\endgroup$
    – jlammy
    Jan 15, 2021 at 16:33
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    $\begingroup$ Here's an example which might help you understand the logic better. Suppose you want to show that if person $X$ and person $Y$ have the same fingerprints, then in fact $X$ and $Y$ are the same person. Then you only care about cases where two people $X$ and $Y$ have the same fingerprints. This is why we "assume" that $X$ and $Y$ have the same fingerprints -- if they don't, then they are irrelevant to the point we are trying to prove. $\endgroup$
    – jlammy
    Jan 15, 2021 at 16:39
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    $\begingroup$ Question $3$: since $a<b$ led to a contradiction, it follows that $a\ge b$ $\endgroup$ Jan 15, 2021 at 16:44
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    $\begingroup$ Yes, @Avra, there's a contradiction in the cases where $a<b$ and $a>b$ $\endgroup$ Jan 15, 2021 at 17:01

1 Answer 1

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This proof really overcomplicates it. Rather, prove the contrapositive:

$$x \neq y \implies f(x) \neq f(y)$$

This is then easy, because either $x < y$ or $y < x$ and thus either $f(x) < f(y)$ or $f(y) < f(x)$. Hence, $f(x) \neq f(y)$ when $x\neq y$.

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  • $\begingroup$ Thank you. I mentioned in the question that this is how I did it, but as you said, the proof above overcomplicates it. $\endgroup$
    – Avv
    Jan 15, 2021 at 16:34
  • $\begingroup$ Can you please summarize in simple words what the proof above is trying to do without fancy math symbols ? $\endgroup$
    – Avv
    Jan 15, 2021 at 16:39
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    $\begingroup$ @Avra Sure. I think in the comments this was already explained just fine? But I would focus on the geometric aspect of the question: is it geometrically obvious to you that a strictly increasing/decreasing function is injective? $\endgroup$
    – J. De Ro
    Jan 15, 2021 at 16:42
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    $\begingroup$ @Avra A ceiling function is not STRICTLY increasing: for example the ceiling function evaluated in $1/3$ and $2/3$ is equal. $\endgroup$
    – J. De Ro
    Jan 15, 2021 at 17:30
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    $\begingroup$ @Avra I don't know either why you got downvoted. I upvoted to compensate this :) $\endgroup$
    – J. De Ro
    Jan 15, 2021 at 22:35

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