1

I have a question concerning the definition of strictly increasing function, that I cannot really figure out. The definition reads:

Definition: A function $f : \mathbb{R} \to \mathbb{R}$ is strictly increasing if $$\forall a, b \ ( a > b \Longrightarrow f(a) > f(b) ).$$

My question is: Why do we have just one conditional and not a biconditional?

That is, why the definition is not the following?

A function $f : \mathbb{R} \to \mathbb{R}$ is strictly increasing if $$\forall a, b \ ( a > b \Longleftrightarrow f(a) > f(b) ).$$

That's how I see it: we don't want to have that $a > b$, and $f(a) \leq f(b)$ (fine with me); but I don't see why we accept to have in principle that $a \leq b$, and $f(a) > f(b)$.

Any feedback is most welcome.
Thank you for your time.

PS: Of course, this is just a particular case of a problem I have with definitions that involve conditionals.

Kolmin
  • 4,083
  • i think you misunderstand the definition – Sathasivam K Oct 10 '16 at 12:49
  • can you just tell me the book from which you got this defintion – Sathasivam K Oct 10 '16 at 12:51
  • Actually, I got it here: http://mathworld.wolfram.com/StrictlyIncreasingFunction.html – Kolmin Oct 10 '16 at 12:54
  • You can see the post : are-if-and-iff-interchangeable-in-definitions. – Mauro ALLEGRANZA Oct 10 '16 at 13:14
  • 1
    Thanks a lot for the link! However, (and I was afraid my way of writing could induce it) there is a misunderstanding: my problem is not with that conditional (the one you emphasized) that I know is actually a biconditional. My problem is with the one immediately after: $$\forall a, b \ ( a > b \Longrightarrow f(a ) > f(b)).$$ That's the conditional I think should be a biconditional. – Kolmin Oct 10 '16 at 13:18
  • @MauroALLEGRANZA: Just wondering, did I make clear my point? Because I know that I am missing something, but at the same time the feedbacks I am getting are sort of out of target. Hence, I still not see what is wrong with my argument. :) – Kolmin Oct 10 '16 at 13:53
  • Now, that's the explanation I was looking for! Very (very) nice. I would love to vote and accept it, if you write it as an answer. Thanks a lot! – Kolmin Oct 10 '16 at 14:38

2 Answers2

1

I think that the trouble is with the "asymmetry" of $<$.

If we use $↔$, we have that the definition of increasing will be:

$f(x) ≤ f(y) ↔ x < y$

and that of strictly increasing will be:

$f(x) < f(y) ↔ x < y$.

We have to consider that $P ↔ Q$ is equivalent to $¬P ↔ ¬Q$, and thus the last one is equivalent to:

$f(x) ≥ f(y) ↔ x ≥ y$.

The case when $x=y$ is not interesting (because for any function : $f(x)=f(y)$, if $x=y$) and thus we have that the new definition of strictly increasing amounts to :

$f(x) ≥ f(y) ↔ x > y$

that coincides with (non-strict) increasing.

Mauro ALLEGRANZA
  • 94,169
  • I can't able to understand your explanation. could you please help me.? – Sathasivam K Oct 22 '16 at 15:13
  • There is a $\geq$ in the first inequality because of the fact that there is a $\geq$ sign in the second inequality. $$f(x) ≥ f(y) ↔ x ≥ y$$ so why do you keep $\geq$ sign in the left condition after removing = sign in the right condition. i disagree with this answer . this doesn't make sense for me.please explain it – Sathasivam K Oct 22 '16 at 15:29
  • For me the converse part should be obviously true. so there is no problem and the statement can be written using bicondition. so if you want to tell that the bicondition does not hold ,then give me a counter examlpe please. – Sathasivam K Oct 22 '16 at 15:49
0

Take for example the function $f(x)=\cfrac {x^2-1}{x-1}$. On every point except $x=1$ it equals $x+1$ hence, it is strictly increasing on the definition.

But, if you take the iff definition you have given: $ 2>1$ means $f(x)$ has to admit: $f(2)>f(1)$ in order to be strictly increasing. which is a meaningless statement.

I'm not saying you can't expand the definition to an "if and only if" definition. You can, but then it will be much longer and in the end it will give the same result as the shorter one.

There is no rule in mathematics against saying Unnecessary things, it's just a waste of time and a headache sometimes.

JonesY
  • 408
  • Thanks a lot for your answer! Still, I am not sure I see your argument, and most probably I did not make clear my point. What I mean is that it seems to me that, as the definition reads now, we allow $f(2) > f(3)$ as a strictly increasing function. – Kolmin Oct 10 '16 at 13:42
  • if say if and only if means for every case of $x_1<x_2$ gives $f(x_1)<f(x_2)$ this is not always a valid argument as shown in my example. You need to be very precise, you can't have an iff definition when the other side have faults. – JonesY Oct 10 '16 at 14:22
  • If your problem is about domain (call it $D$), codomain (call it $C$) and definability, we can rephrase things as follows: the (standard) definition reads $$\forall a, b \in D\ (a > b \Longrightarrow f(a) > f(b) ),$$ where I just add the other side $$\forall a, b \in D \ (f(a) > f(b) \Longrightarrow a > b),$$ and I don't see your point anymore. – Kolmin Oct 10 '16 at 14:34
  • @Kolmin, You asked, when given an "if ... then..." definition, why not to make it always an "if and only if" definition. My answer is that sometimes it is not easy to make an iff definition. Even when it is easy to make the iff definition, unless you have to, why should you waste time? There is no law against saying unnecessary things in math, no one will kill you if you take every "if... then..." definition at make it an iff one successfully, it's just a waste of time to do it sometimes. – JonesY Oct 11 '16 at 13:41
  • I understood your point, but I found inconclusive and slightly illposed in this context. On the contrary, MAURO's answer actually answers the question. The point is that your implicit reference to the Occam's razor – IMO – is not to the point. True, you don't want to multiplicate objects when unnecessary, but here it's not about that: rather it is about understanding a definition, which is by force something already established. There is nothing really philosophical here, it is all purely technical. I hope this clarifies my point. – Kolmin Oct 11 '16 at 15:22