# Prove that a strictly decreasing function from $f:\Bbb R \to \Bbb R$ is one-to-one

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 $$b$$.
• 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. Jan 15, 2021 at 16:29
• 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. Jan 15, 2021 at 16:33
• 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. Jan 15, 2021 at 16:39
• Question $3$: since $a<b$ led to a contradiction, it follows that $a\ge b$ Jan 15, 2021 at 16:44
• Yes, @Avra, there's a contradiction in the cases where $a<b$ and $a>b$ Jan 15, 2021 at 17:01

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 $$x\neq y$$.

• Thank you. I mentioned in the question that this is how I did it, but as you said, the proof above overcomplicates it.
– Avv
Jan 15, 2021 at 16:34
• Can you please summarize in simple words what the proof above is trying to do without fancy math symbols ?
– Avv
Jan 15, 2021 at 16:39
• @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? Jan 15, 2021 at 16:42
• @Avra A ceiling function is not STRICTLY increasing: for example the ceiling function evaluated in $1/3$ and $2/3$ is equal. Jan 15, 2021 at 17:30
• @Avra I don't know either why you got downvoted. I upvoted to compensate this :) Jan 15, 2021 at 22:35