Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to prove the statement in the title.

Proof: We prove that if $f$ is not strictly decreasing, then it must be strictly increasing. So suppose $x < y$.

And that's pretty much how far I got. Help will be appreciated.

share|cite|improve this question
up vote 7 down vote accepted

Prove the contrapositive instead: if $f$ is not strictly increasing and not strictly decreasing, then it is not-one-ot-one.

For example, say there are points $a\lt b\lt c$ such that $f(a)\lt f(b)$ and $f(b)\gt f(c)$. Either $f(a)=f(c)$ (in which case $f$ is not one-to-one), or $f(a)\lt f(c)$, or $f(c)\lt f(a)$.

If $f(a)\lt f(c)\lt f(b)$, then by the Intermediate Value Theorem there exists $d\in (a,b)$ such that $f(d)=f(c)$; hence $f$ is not one-to-one.

Now, there are other possibilities (I made certain assumptions along the way, and you should check what the alternatives are if they are not met).

share|cite|improve this answer
Thanks, I knew you went for the kill with the IVT; just wasn't sure how to set it up. – WacDonald's Jul 13 '12 at 3:19

Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous and not strictly increasing. Then there exists two points such that $f(a) = f(b)$, or there exists three points $a < b < c$ such that $f(a) < f(b)$ and $f(b) < f(c)$. The first case contradicts injectivity. Suppose the second, without loss of generality, suppose that $f(b) - f(a) \leq f(c) - f(b)$. Then $f(b) \leq f(b) - (f(b) - f(a)) = f(a) \leq f(c)$. By the intermediate value theorem, there exists $d$ such that $b < d < c$ such that $f(d)= f(a)$. This contradicts injectivity.

share|cite|improve this answer

Since $f$ is one-to-one, $f(a)\neq f(b)$. We first consider the case when $f(a)<f(b)$. I claim that in this case $f$ is strictly increasing.

First, note that for any $x\in(a,b), f(a)<f(x)$. If not, then since $f$ is 1-1, must have $f(x)<f(a)$. But then by the IVT there is some $c\in(x,b)$ so that $f(x)=f(a)$, contradicting $f$ being 1-1.

Now suppose for contradiction that $f$ is not strictly increasing. So there is some $x,y\in I$ with $f(y)\leq f(x)$. However, since f is 1-1, we cannot have $f(y)=f(x)$, so $f(y)<f(x)$. By the previous paragraph, we also have $f(a),f(x)$. So by the IVT, there is some $c\in (a,x)$ with $f(c)=f(y)$, contradicting $f$ being 1-1.

Thus $f$ is strictly increasing if $f(a)<f(b)$. If $f(b)<f(a)$, a similar argument (with all inequalities reversed) shows $f$ is strictly decreasing.

share|cite|improve this answer
The first sentence makes no sense. In any event, correct answers were given over a year ago, and I'm not sure how you propose to improve on them. – user43208 Oct 24 '13 at 19:26
@user43208 Which part of the first sentence doesn't make sense to you? – user87274 Oct 24 '13 at 19:30
To say $f$ is one-to-one means distinct arguments $a, b$ are carried to distinct values $f(a), f(b)$. So it makes no sense to conclude $f(a) = f(b)$ (which you then proceed to gainsay in the very next sentence). – user43208 Oct 24 '13 at 19:33
Oh my mistake. It should be $f(a)\neq f(b)$. – user87274 Oct 24 '13 at 19:39

Consider $g\colon \{(x,y)\mid x<y\}\mapsto\mathbb R$, defined by $g(x,y):=f(x)-f(y)$. Clearly $g$ is continuos. Since the domain of $g$ is connected and $g$ has no zeroes, the image of $g$ is an interval not containing $0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.