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A lot of the time in lectures, my professors prove (by induction) an inequality (e.g. $(1+x)^n \geq 1+nx$) in the natural numbers (or any subsets thereof), and I've noticed (not rigourously; only by graphing the functions) that such statements are also true for all real numbers inbetween.

Another example is that exponential growth beats polynomial growth.

My question is:

If an inequality is true for all $n \in \mathbb{N},$ does it necessarily follow that the same inequality is true for all $n \in \mathbb{R^+}$?

I'm not in the market for a rigourous proof; (if the answer's no) just a counter-example, or (if the answer's yes) an intuitive reason why this is the case.

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    $\begingroup$ I don't think that's necessarially true, but I know that if it's true for $\mathbb{Q}$ then it's true for $\mathbb{R}$. Since any real number can be expressed as a limit of a sequence over the rationals. $\endgroup$
    – Darth Geek
    Jul 30, 2014 at 10:46
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    $\begingroup$ @DarthGeek If, of course, continuous functions are involved. Without continuity, you have nothing. $\endgroup$
    – 5xum
    Jul 30, 2014 at 10:47
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    $\begingroup$ @DarthGeek the indicator function of the rationals is a counterexample $\endgroup$
    – Mathmo123
    Jul 30, 2014 at 10:47
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    $\begingroup$ @DarthGeek additionally, the limiting process only necessarily preserves weak inequalities (i.e. $\leq$ or $\geq$) but not strict ones, so you'd have to say more to prove the claim. $\endgroup$
    – Hayden
    Jul 30, 2014 at 10:47
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    $\begingroup$ BTW, $\forall n\in\mathbb N_0,\forall x\in[-1,\infty)\colon (1+x)^n\ge 1+nx$ (note that one of the quantifiers is already about real numbers!) is already a counterexample: With $n=\frac12$ we do not have $\sqrt{1+x}\ge1+\frac12x$ for all$x\ge -1$. For example $\sqrt{1+48}=7<25=1+\frac{48}2$. $\endgroup$
    – Hagen von Eitzen
    Jul 30, 2014 at 11:08

6 Answers 6

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$x - \lfloor x \rfloor\le 0$, (where $\lfloor x \rfloor$ is the largest integer less than or equal to $x$) is true for every $x \in \mathbb N$ but false for all other positive real numbers.

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  • $\begingroup$ Great example, thanks for sharing $\endgroup$
    – TheNotMe
    Jul 30, 2014 at 12:06
  • $\begingroup$ This answer is wrong, since $\;x - \lfloor x \rfloor\le 0 \;$ holds for $\; x = -1 \:\:$. $\;\;\;\;\;$ $\endgroup$
    – user57159
    Jul 30, 2014 at 21:58
  • $\begingroup$ The question was about positive reals. This answer is easily remedied to exclude negative integers too - eg take $|x|-\lfloor x \rfloor \le 0$ $\endgroup$
    – Mathmo123
    Jul 30, 2014 at 22:02
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    $\begingroup$ @RickyDemer $\mathbb N$ usually means $\{x: x \ge 0,\; x\text{ is an integer}\}$. $\endgroup$
    – michaelb958--GoFundMonica
    Jul 30, 2014 at 23:10
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Here is a more precise version of your question:

Question. If a statement of the form $\tau \geq \sigma$ (where $\tau$ and $\sigma$ are expressions built using only the operations $\{0,1,+,\times\}$) holds for $\mathbb{N},$ does it necessarily hold for $\mathbb{R}_{\geq 0}$?

Unfortunately, the answer is no, (thanks @Mathmo123).

Consider $x^2 \geq x$. This holds for all $n \in \mathbb{N}$ (in fact, for all integers), but not for the $r \in \mathbb{R}$ strictly between $0$ and $1$. Of course, a counterexample is rigorous proof of falsity.

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    $\begingroup$ $x^2 \ge x$ is an example $\endgroup$
    – Mathmo123
    Jul 30, 2014 at 10:58
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    $\begingroup$ A strict version would be $5x^2+1>5x$ $\endgroup$
    – Hagen von Eitzen
    Jul 30, 2014 at 11:04
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    $\begingroup$ By solving the quadratic and noticing its roots are between consecutive integers. $\endgroup$
    – zibadawa timmy
    Jul 30, 2014 at 11:36
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    $\begingroup$ I like the example above as it's continious. $\endgroup$
    – Dmitry Ginzburg
    Jul 30, 2014 at 12:00
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    $\begingroup$ I didn't think to deeply about it, but what I thought was going on is this: the expressions allowed as $\sigma$ and $\tau$ in the more precise questions are polynomials of one real variable. Their difference is also a polynomial, and $\sigma \leq \tau$ if and only if this new polynomial $\sigma - \tau$ is less than or equal to zero. Being a polynomial its graph can cross the $x$-axis only a finite number of times. So for numbers bigger than some $R$ it must be all negative or all positive and evaluating at an integer tells you (by assumption) that it must be all negative. Similarly for $< L$. $\endgroup$
    – Vincent
    Aug 1, 2014 at 12:49
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Cosine of $2n\pi$ is greater than zero for every integer $n$, but not for every real $n$...

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Without further requirements on the inequality, the answer is a massive no, as all respondent say. This is because there is not constraint at all between neighboring values of the argument.

The answer would be different and much more interesting if you imposed smoothness conditions, like continuity to some order, differentiability to some order, Lipschitz continuity, band-limitedness...

Your question is probably hiding a deeper one: given a discrete function, is there a "natural" way to define an extension to real values ? And what properties would it possess ?

A good example of such an extension is given by the Gamma function, $\Gamma(x+1)$, that generalizes the factorial $n!$.

If your inequality is $n!\ge1$, it is violated between $0!$ and $1!$, but this is quite understandable as the two first values form a plateau.

http://en.wikipedia.org/wiki/File:Factorial_Interpolation.svg

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$\sin(n\pi) \geq 0$ for all integers $n$...

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$n > 0.5$ is true for all $n \in N$.

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    $\begingroup$ ...except for 0. Which is excluded from some definitions of the natural numbers for murky reasons, but the question seems to assume it's included. (i.e., equating the positive reals with the ones "between the naturals") $\endgroup$
    – camccann
    Jul 31, 2014 at 2:41
  • $\begingroup$ The question speaks of real numbers "in between", and I indeed consider it more reasonable to deal with interpolation than extrapolation. In the case of the inequality $>0.5$, the property holds "in between". $\endgroup$
    – user65203
    Jul 31, 2014 at 17:04

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