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I have a question about infinity. (I'm in 8th grade so please just let me know if this is a stupid question, and please ask if you want any clarification).

How do you define an infinitely small value that is greater than zero?

$1/\infty$ is apparently invalid because you can't divide by infinity. And $0.\bar{0}1$ is invalid because you can't put a 1 after an infinite number of zeros.

So how do use infinitely small values in equations? If I say $x$ tends to $0$ and $ x > 0$ then is x an infinitely small positive value?

If $x \rightarrow 0$; $ x > 0$ is invalid, how else could I define an infinitely small value? And if I could define $x$ as being in infinitely small value, could I then say that $ x * \infty = 1 $?

And, is there a symbol for an infinitesimal (infinitely small) positive value?

Thank you.

EDIT:

I realize that for every number there is a smaller one (e.g. $x/2$) so therefore there is no "smallest" number - I'm looking for a symbol (or equation/name/concept) that stands for any (for lack of a better word) "number" infinitely close to zero. Similar to the way in which infinity isn't really a number, but stands for any "number" infinitely large. I know $\infty$ isn't a number! That's why I put "quotes" around the word "number"!

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  • $\begingroup$ i think they generale use either $\epsilon$ or $\delta$ for infinetesimals $\endgroup$ – cand Mar 6 at 4:31
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    $\begingroup$ I'll leave this here as a comment, as I'm sure someone will give a more detailed response. But for further reading, the archimedian principle may be of interest: en.m.wikipedia.org/wiki/Archimedean_property $\endgroup$ – Ryan Goulden Mar 6 at 4:32
  • $\begingroup$ Related: math.stackexchange.com/questions/455639/… $\endgroup$ – d.k.o. Mar 6 at 4:32
  • $\begingroup$ It, in short, says that there are no numbers (or perhaps, in your terminology, "values") which are either infinitely large, or infinitely small. $\endgroup$ – Ryan Goulden Mar 6 at 4:33
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    $\begingroup$ And it is absolutely a good question! These are the kinds of philisophical intuitions used in the development of analysis. Certainly good to be thinking about! $\endgroup$ – Ryan Goulden Mar 6 at 4:35
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The thing is $\infty$ doesn't refer a number that is "infinitely big." There is no such thing, nor does it mean a number that is arbitrarily large -- such a number would still be finite -- nor does it refer to any actual values.

It is somewhat incorrect to say things like $1+1+1+.... = \infty$ or $\frac 10 = \infty$ or $\frac 1\infty = 0$. Those are all incorrect statements.

More technically the correct way to state those would be as limits:

$\lim_{n\to \infty} \underbrace{1+1+1+.... +1}_{n\text{ times}} = \infty$, or $\lim_{x\to 0^+} \frac 1x = 0$ and $\lim_{x\to \infty} \frac 1x = 0$.

The phrase "$\lim_{x\to c} somethingexpressedintermsof(x) =k$" means.... well, informally, it means, "as $x$ gets closer and closer to $c$ then $somethingexpressedintermsof(x)$ gets closer and closer to $k$".

For example $\lim_{x\to 5} x^2 = 25$.

Now if instead of getting "closer and closer" to something specific, we want to express the idea that the values get "larger and larger" and that, if you imagine any number as larger as you want this value will at some point get that large and even larger than that -- that's when we use the symbol $\infty$.

BUT THIS IS IMPORTANT! The symbol doesn't represent any actual value. It only represents the idea that a series of values can get larger and for any large value you can imagine the series of values will somewhere get as large as that.

So $\lim_{x\to \infty} \frac 1x = 0$ means if we take $x = 1,000$ then $\frac 1x$ will be close to $0$ and if we take a larger value like $x= 10^{1,000}$ or $x=10^{googol}$ or $x$ getting "larger and larger" we get $\frac 1x$ gets larger and larger.

Alternatively $\infty$ can be in the "detonation spot". $\lim_{x\to 5}\frac 1{x-5} = \infty$ means this: if we take $x$ really really close to $5,$ then $\frac 1{x-5}$ will be really really large. And we can make $\frac 1{x-5}$ as large as we like by taking $x$ as close to $5$ as we like.

.....

Okay, so if $\infty$ is the symbol for getting things really really large what is the symbol for getting things really really small. Well, ... that means "getting close to $0$". We had that all the time.

$\lim_{x\to 0} \frac {x^2 + 4x}{x}=4$ means "if we take $x$ to be as arbitrarily small as we like". And that's pretty much the concept you want. There's no such thing as $\infty$ or a number that is "infinitely large" and there is no such thing as a number that is "infinitely small". And if a number getting infinitely large "tends toward infinity", the number getting infinitely small will "tend toward zero".

And that's pretty much it.

... HOWEVER...

We want to be able to do math on these small values -- these values that are arbitrarily close to zero but not quite at zero.

When we do that we frequently say: "Let $\epsilon > 0$ be arbitrarily small" Or "Let $\delta$ be an arbitrarily small positive number" or "let $h > 0$". In these cases though, it's important to realize that although we are trying to figure out what happens when these variables get "infinitely small", at no time or value are these numbers anything strange or different. These variables represent actual normal numbers. Just very small ones.

For example, consider this problem: if $f(x) = x^2 + 5,$ what is $\lim_{x\to 5} \frac {f(x) -f(5)}{x-5}$?

Let $|h| > 0$ be arbitrarily small. Then if $x = 5 + h$ we have

$\frac {f(5+h) - f(5)}{(5+h) - 5} = \frac {(25 + 10h + h^2) - 25}{h}= \frac {10h + h^2}{h} = 10 + h$.

So $\lim_{x\to 5} \frac {f(x) -f(5)}{x-5}$

$= \lim_{h\to 0} \frac {f(5+h) - f(5)}h $

$ = \lim_{h\to 0}10 + h = 10 + 0 = 10$.

Notice that although $h$ is our "infinitely small" number, it is an actual number.

Also note that if $h$ does actually equal $0$ we do not get $\frac {f(5+h) - f(5)}h = 10$. We get $\frac {f(5+h) - f(5)}h=\frac 00$ is undefined garbage. This only works if $h$ is "arbitrarily small" and we get $\frac {f(5+h)-f(5)}h = 10 + h,$ and as $h$ is "arbitrarily small" this $\to 10$ as $h\to 0$.

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  • $\begingroup$ Mmm.... kinda.... there's a bit more finesse than that. $\endgroup$ – fleablood Mar 6 at 18:40
  • $\begingroup$ I think you meant $1 \over x$ gets smaller and smaller. $\endgroup$ – J. W. Tanner Mar 6 at 22:54
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The short answer is that you can't. But you can define any positive value which is arbitrarily small, in other words as small as you want it to be.

This means that just like how there is no "largest number" (since for any number $N$, we know that $N + 1$ is larger), there is no smallest positive number, since for any real number $N \in \mathbb{R}$,

$$\frac{1}{N} > \frac{1}{N+1} $$

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    $\begingroup$ Or indeed, no matter how small the positive number $a$ is, $a/2$ is smaller. $\endgroup$ – Lubin Mar 6 at 5:19
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    $\begingroup$ @Lubin Your example is clearer, thanks! $\endgroup$ – Sean Lee Mar 6 at 5:20
  • $\begingroup$ But, as you said, arbitrarily small isn't small enough if I want something to be "infinitely small". $\endgroup$ – scitronboy Mar 6 at 17:01
  • $\begingroup$ Something interesting to add here is that there is nothing special about not being able to make an infinitely small number. You also can make a number that is infinitely closest to 1, nor infinitely close to 10, or any number. $\endgroup$ – Prince M Mar 6 at 17:19
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    $\begingroup$ Also, what do we mean by "infinitely close". Many might take "infinitely close to 1" as being the same thing as "equalling 1". In fact, many mathematicians in the past were content invoking "what is true up to the limit, is true at the limit", which of course, wouldn't suffice by todays standards of proof. $\endgroup$ – Ryan Goulden Mar 6 at 17:36
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There is a simple, concrete way to do this. Consider the set $\mathbb{R}(x)$ of all rational functions of a real variable, $$ f(x) = \frac{p(x)}{q(x)}, $$ where $p(x)$ and $q(x)$ are polynomials in $x$. [I'm not going to worry about the distinction between 'polynomials' and 'polynomial functions', nor shall I worry about the fact that the denominator of $f(x)$ may be zero for a finite number of values of $x$ - these worries can be taken care of.]

$\mathbb{R}(x)$ includes the set of all constant functions $f(x) = t$, where $t$ is any real number, in particular the zero function $f(x) = 0$. Rational functions can be added, subtracted, multiplied, and divided (except that of course you can't divide by the zero function); these operations obey the usual algebraic laws; and their results coincide with the results of the usual operations in the case of the constant functions. We say that $\mathbb{R}(x)$ is a field, which extends the real field $\mathbb{R}$. [All of this can be made quite formal and watertight.] In practice, we make no distinction between elements of $\mathbb{R}$ and the corresponding 'constant' elements of $\mathbb{R}(x)$. [I don't think that this practice is ever actually made formal - but again let's just not worry about it now!]

Although $x$ here is a 'variable' and not a 'number' [and it is important to make this distinction when one defines polynomials properly - especially when dealing with fields other than $\mathbb{R}$], there is a clear sense in which $x$ can be regarded as an infinitesimally small 'number'. (There is also a similar but different treatment which makes $x$ an infinitely large 'number'.) The point is that (i) the field $\mathbb{R}(x)$ can be totally ordered, in a way that extends the usual ordering of $\mathbb{R}$, and (ii) in this ordering of $\mathbb{R}(x)$, we have $0 < x < t$ for every positive real number $t$.

The definition of the total ordering of the field $\mathbb{R}(x)$ is straightforward. [Caution: I haven't written out the details before, so it is a good and necessary exercise to check that I haven't slipped up! Also, I don't have an authoritative reference to hand at the moment. The Web page A non Archimedean ordered field is helpfully clear, but unfortunately it is also erroneous - it gets confused, at the end, between the two constructions in which $x$ may be either infinitesimal or infinite.] Say that a non-zero rational function, $$ f(x) = \frac{p(x)}{q(x)} = \frac{a_nx^n + a_{n+1}x^{n+1} + a_{n+2}x^{n+2} + \cdots}{b_mx^m + b_{m+1}x^{m+1} + b_{m+2}x^{m+2} + \cdots}, $$ where $a_n \ne 0$ and $b_m \ne 0$, is positive if $a_n$ and $b_m$ have the same sign.

(Intuitively, if $x$ is small, then all powers of $x$ but the lowest are negligible, so $f(x) \bumpeq a_nb_m^{-1}x^{n-m}$. If $x$ is positive, it follows that $f(x)$ is positive if and only if $a_n$ and $b_m$ have the same sign.)

This definition is independent of the particular expression of $f(x)$ as a ratio of polynomials, because if also $$ f(x) = \frac{r(x)}{s(x)} = \frac{c_kx^k + c_{k+1}x^{k+1} + c_{k+2}x^{k+2} + \cdots}{d_lx^l + d_{l+1}x^{l+1} + d_{l+2}x^{l+2} + \cdots}, $$ where $c_k \ne 0$ and $d_l \ne 0$, then $p(x)s(x) = q(x)r(x)$, whence $a_nd_l = b_mc_k$, from which it is clear that $a_n$ and $b_m$ have the same sign if and only if $c_k$ and $d_l$ have the same sign.

For rational functions $f(x)$ and $g(x)$, we now write $f(x) > g(x)$ if and only if $f(x) - g(x)$ is positive. [The symbols $\geqslant$, $<$ and$\leqslant$ may then defined in terms of $>$, for $\mathbb{R}(x)$, in the usual way.]

In $p(x)/q(x)$, because $b_m \ne 0$, we can divide $p(x)$ and $q(x)$ by $b_m$, and thus can assume without loss of generality that the trailing coefficient of $q(x)$ is $1$. Then $p(x)/q(x)$ is positive if and only if $a_n > 0$.

Proposition. If $f(x)$ and $g(x)$ are positive, then so are $f(x) + g(x)$ and $f(x)g(x)$.

Proof. Let $f(x) = p(x)/q(x)$, and $g(x) = r(x)/s(x)$, where the polynomials $p(x)$, $q(x)$, $r(x)$ and $s(x)$ are as above, and $b_m = d_l = 1$. Suppose $f(x)$ and $g(x)$ are positive, i.e. $a_n > 0$ and $c_k > 0$. $$ f(x)g(x) = \frac{p(x)r(x)}{q(x)s(x)}. $$ The trailing coefficient of $p(x)r(x)$ is $a_nc_k > 0$, and the trailing coefficient of $q(x)s(x)$ is $b_md_l = 1$, therefore $f(x)g(x)$ is positive. $$ f(x) + g(x) = \frac{p(x)s(x) + q(x)r(x)}{q(x)s(x)}. $$ Again, the trailing coefficient of $q(x)s(x)$ is $b_md_l = 1$. The trailing coefficient of $p(x)s(x)$ is $a_nd_l = a_n$, and the trailing coefficient of $q(x)r(x)$ is $b_mc_k = c_k$. Therefore, the trailing coefficient of $p(x)s(x) + q(x)r(x)$ is $a_n$, $c_k$, or $a_n + c_k$, according as $n+l < m+k$, $n+l > m+k$, or $n+l = m+k$, respectively. In all three cases, the coefficient is positive, therefore $f(x) + g(x)$ is positive. $\square$

["$\square$", or a similar symbol, is often used to mark the end of a proof.]

This proposition enables us to operate just as confidently with the order properties of $\mathbb{R}(x)$ as with its arithmetic (or perhaps I should say, algebraic) properties. Elements of $\mathbb{R}(x)$ thus behave far more like 'numbers' than one would at first expect. $\mathbb{R}(x)$ certainly extends the familiar structure of $\mathbb{R}$ in terms of its total ordering, as well as in terms of addition, subtraction, multiplication, and division - even if one chooses not to regard the non-constant elements of $\mathbb{R}(x)$ as 'numbers'.

As promised, some of these new 'numbers', including our old friend $x$, are infinitesimally small: if $t$ is any positive real number, no matter how small, then $$ t - x = \frac{t - x}{1} > 0, $$ simply because $t$ and $1$ have the same sign. Similarly, the rational function $\frac{1}{x}$ is infinitely large: for all real $t$, $$ \frac{1}{x} - t = \frac{1 - tx}{x} > 0, $$ because the coefficients of the lowest powers of $x$ in the numerator and denominator are both $1$.

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  • $\begingroup$ R. Hartshorne, Geometry: Euclid and Beyond (2000), Prop. 18.1 briefly describes the construction in which $x$ is larger than any real number. Similar brief descriptions can be found on the Net (including MSE); but the construction in which $x$ is smaller than any real number only ever seems to get mentioned in passing (if at all). T. S. Blyth, Lattices and Ordered Algebraic Structures (2005), Theorem 10.4 carefully constructs the totally ordered field of quotients of a totally ordered integral domain, but doesn't mention either of the total orderings of the polynomial ring $\mathbb{R}[x]$. $\endgroup$ – Calum Gilhooley Mar 6 at 22:40

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