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This question already has an answer here:

I am trying to describe how irrational numbers, which are all modeled as a series of fractions, can themselves not be fractions, and are instead part of a unique group of "decimal numbers" outside of fractions, called the irrational numbers. I am confused atm.


From Wikipedia, some example irrational numbers include:

  • $\sqrt 2$
  • the golden ratio
  • The sqrt of all natural numbers which are not perfect squares
  • Logarithms

Then they say:

Almost all irrational numbers are transcendental and all real transcendental numbers are irrational. Examples include $e^\pi$.

Rational numbers are fractions, which are included in the set of irrational numbers. Irrational numbers, however, are decimals and include things that "can't be represented as fractions" it seems.

But where I'm confused is, sqrt 2 can be represented by a series of fractions:

$${\displaystyle {\sqrt {2}}=\prod _{k=0}^{\infty }{\frac {(4k+2)^{2}}{(4k+1)(4k+3)}}=\left({\frac {2\cdot 2}{1\cdot 3}}\right)\left({\frac {6\cdot 6}{5\cdot 7}}\right)\left({\frac {10\cdot 10}{9\cdot 11}}\right)\left({\frac {14\cdot 14}{13\cdot 15}}\right)\cdots }$$

Similarly, $\pi$ can be represented by a series of fractions:

$${\displaystyle 1\,-\,{\frac {1}{3}}\,+\,{\frac {1}{5}}\,-\,{\frac {1}{7}}\,+\,{\frac {1}{9}}\,-\,\cdots \,=\,{\frac {\pi }{4}}.}$$

Finally, the natural logarithm can be written as a series of fractions:

$${\displaystyle \ln(1+x)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}x^{k}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots }$$

It has been a while since I have added/divided/subtracted/multiplied fractions, but from what I remember doing any of those operations results in a new fraction. So I'm wondering what I'm missing when it comes to understanding irrational numbers. If irrational numbers can represent non-fraction numbers, yet they are themselves represented by a series of fractions, it seems the result of the series would itself be a fraction, and so the irrational numbers are all rational numbers. Looking for an understanding of how to explain the difference between rational and irrational numbers. I tried saying "irrational are decimal numbers you can't represent with a fraction", but then when getting into the definition of a rational numbers (fraction numbers), I was unable to explain how if all irrational numbers are themselves definable as a series of fractions, how they themselves aren't representable as fractions. Thank you for your help.

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marked as duplicate by Hans Lundmark, Carsten S, Matthew Towers, Lord Shark the Unknown, Namaste Jun 20 '18 at 22:04

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    $\begingroup$ Rational numbers are fractions A fraction is the ratio of two integers. sqrt 2 can be represented by a series of fractions An infinite series of fractions is not necessarily a fraction itself. $\endgroup$ – dxiv Jun 20 '18 at 3:19
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    $\begingroup$ You say Rational numbers are fractions, which are included in the set of irrational numbers. Irrational numbers, however, are decimals, and include things that "can't be represented as fractions" it seems. This is incorrect. The set of real numbers is the union of the sets of rational and irrational numbers. No "fraction" (rational number) is an irrational number. Indeed, an irrational number is, by definition, a number which cannot be expressed as the ratio of two integers (ratioal numbers). $\endgroup$ – Xander Henderson Jun 20 '18 at 3:23
  • $\begingroup$ Related: What is a real-world metaphor for irrational numbers?. $\endgroup$ – Rodrigo de Azevedo Jun 20 '18 at 7:04
  • $\begingroup$ Adding, subtracting, multiplying, or dividing TWO rational numbers yields a rational number, and consequently so it does with more than two, as long as there are finitely many. Thus $A+B+C+D$ is a sum of TWO numbers $A+B$ and $C+D,$ and each of those is a sum of two numbers, and you can do something similar with $A+B+C+\cdots+Z,$ and so on. $\endgroup$ – Michael Hardy Jun 20 '18 at 16:33
  • $\begingroup$ Re, "logarithms": Every real number is a logarithm of some positive real number. Some real numbers are rational numbers. Therefore, some logarithms are rational numbers. $\endgroup$ – Solomon Slow Jun 20 '18 at 17:06
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Rational numbers are values the can be written as a ratio (fraction) of two whole numbers. Irrationals are those values that can not.

End of story.

The first trick is to realize that there are values that can't be (Pythagoras didn't want to believe it). But if you toss a dart at something a mile wide what is the likelihood that it will hit a value that is exactly a ratio of two whole numbers. If one puts it that way it seems slim.

But the second trick is to wonder, if the value isn't a ratio of whole numbers, then what is it and how can we express it? The hardest thing for novices to get is that question doesn't have an answer. We can't express them.

But fractions can be made to be at least as small and smaller than anything we like. That means although we can not express all values, we can express a rational fraction that can be at least and closer to the value as we'd like.

Now decimals are just fractions. $0.1=\frac 1 {10} $ and $0.3769=\frac {3769}{10000} $. We can get as close to expressing all values by taking longer and longer decimals. But we can never actually express an irrational with decimals. Instead ever Irrational has an infinite string of decimals that get close to it to any possible degree.

And that is what we mean be an irrational number being a series of rational fractions. For every irrational number we can find a series (not just one; many series- the decimal expression is one of them) of rational numbers getting closer and closer together and honing in on the irrational and not honing in on anything else. The series is infinite and if we were immortal gods existing out of time and space we could see the entire infinite series at once and see it converges precisely to the irrational numbers. But we are not immortal gods outside of space so we can not. But we know it does.

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    $\begingroup$ Certain numbers can be expressed as fractions -- meaning they are rational numbers -- but cannot be precisely expressed as decimal -- e.g. 1/3 or 1/7. Is there any reason to introduce the topic of decimals here? $\endgroup$ – Zev Spitz Jun 20 '18 at 10:39
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    $\begingroup$ While it is true that the uniform probability of choosing a rational number from an interval is 0, if you throw a dart it will certainly touch many rational numbers. :-) $\endgroup$ – Pablo H Jun 20 '18 at 14:53
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    $\begingroup$ I felt a little guilty about the dart example. It's basically "white lie we tell children". The thing is, if we teach numbers as we tend to, we teach first whole numbers, then rational fractions, then decimals, and that an irrational is a number that can't be a whole number fraction and thus has infinite non-repeating decimal. This is a bad way to teach IMO and the most students (not insensibly) have a difficult time concieving how this make sense and can't understand how a number CAN'T be rational.... to be continued... $\endgroup$ – fleablood Jun 20 '18 at 16:29
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    $\begingroup$ Re, "If you toss a dart at something a mile wide,..." why would the width matter? There is no limit to how large the denominator of a fraction can be. There is a countably-infinite number of rational number in between any two points on the number line. $\endgroup$ – Solomon Slow Jun 20 '18 at 16:50
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    $\begingroup$ @Barmar, I'll have to argue that the large size just makes it easier to misunderstand what "irrational" really means. $\endgroup$ – Solomon Slow Jun 20 '18 at 17:26
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Let's start by pinning down the definition of "(ir)rational" precisely, since it's sometimes presented unclearly. A rational number is a ratio of integers: $a$ is rational if there are integers $x, y$ such that $a={x\over y}$. Every number can be written as a fraction, it's the "of integers" part which is crucial. Certainly every number with a terminating (= finitely long) decimal expansion is rational: to see why, think about the (completely arbitrary) example $$254.1345987={2541345987\over 10000000}.$$ (Note how much clearer this is than "a rational number is a fraction"!)

Moving on to the other part of your question: just because a number is a "limit" (which can be made precise!) of rational numbers doesn't mean that it itself is a rational number. In fact every number is a limit of rational numbers (that is, the rationals are dense in the reals) in a very simple way: just consider increasingly long decimal representations! E.g. $\pi$ is the limit of the sequence $$3, {31\over 10}, {314\over 100}, {3141\over 10000},...$$

In general, just because $a$ is a limit of things with property $P$ doesn't mean that $a$ itself has property $P$.

EDIT: At this point it might help to see an explicit example of a "naive limit argument" breaking down. Let's think about terminating decimal expansions. It's easy to show that a number $a$ has a terminating decimal expansion iff $a$ can be written as $p\over q$ where $p, q$ are integers and $q$ is a power of $10$. For example, ${3\over 50}$ fits the bill since ${3\over 50}={6\over 100}$ and $100$ is a power of $10$.

Now think about the number $1\over 3$. This doesn't have a terminating decimal expansion - its (unique) decimal expansion is $0.33333...$. However, we can write it as $${1\over 3}=0.3+0.03+0.003+0.0003+...$$ and $0.3, 0.03, 0.003, 0.0003, ...$ all have terminating decimal expansions. Alternatively, we could write ${1\over 3}$ as the limit (again, in a rigorous sense which I won't define here) of the sequence $$0.3, 0.33, 0.333, 0.3333, ...$$ and each term of this sequence has a terminating decimal expansion. So we can see that even though ${1\over 3}$ can be "approached by" or "built from" numbers with terminating decimal expansions, it itself does not have a terminating decimal expansion.

(Yet another example: $0$ is a limit of positive numbers, but is not positive. And so on.)


Note that none of this addresses the question of how we know that there are irrational numbers; that's a separate problem. There are some numbers which we can rigorously prove are irrational, and more abstractly there is a technique in set theory which shows that most numbers are irrational (and most numbers are transcendental, and ...) in a precise sense. But that's a separate issue.

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  • $\begingroup$ This is interesting yet I am still unable to explain how if fractions like sqrt 2 is the result of a series of fractions, how it itself is not a fraction. I would like to be able to explain this in non-mathematical terms if possible. Also not sure what is meant here: "just because $a$ is a limit of things with property $P$ doesn't mean that a itself has property $P$" $\endgroup$ – Lance Pollard Jun 20 '18 at 3:24
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    $\begingroup$ @LancePollard It's the other way around, really: why should it be? The onus is on you (taken broadly) to explain why just being a limit of rationals is enough for something to actually be rational. $\endgroup$ – Noah Schweber Jun 20 '18 at 3:26
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    $\begingroup$ @LancePollard "just because $a$ is a limit of things with property $P$ doesn't mean that $a$ itself has property $P$" This is just the abstract (negative) principle at work here. For example, just because a number $a$ is a limit of rational numbers doesn't mean that $a$ is rational (here "property $P$" is "is rational"). My point is that the jump from "it's a limit of things with [property]" to "ok, it itself actually has [property]" is unwarranted (without further argument), and indeed false in general. $\endgroup$ – Noah Schweber Jun 20 '18 at 3:28
  • $\begingroup$ Ok thank you, will have to learn more about that. $\endgroup$ – Lance Pollard Jun 20 '18 at 3:31
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    $\begingroup$ @LancePollard It's worth mentioning that the general question of when/to what extent "limit arguments" are valid is a (the?) focus of real analysis (more or less "foundations of calculus," although that's a somewhat unfair gloss). $\endgroup$ – Noah Schweber Jun 20 '18 at 3:44
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Any finite sum of rational numbers is rational. Even if it is a sum of millions or quadrillions of rational numbers. I think the step which is hardest to comprehend is understanding the limit that finally leads from only rational numbers to the irrational number.

However, what does it even mean to add infinitely many rational numbers? This is not clear a priori and needs a definition. Let's see how to motivate one.

Some rational sums seem to tend to another rational:

$$1+\frac12+\frac14+\frac18+\cdots.$$

the partial sums $1$ and $1+1/2=1.5$ and $1+1/2+1/4=1.75$ and so on come closer to $2$ from below but never exceed it. It was therefore defined that this sequence converges to the number $2$. This is merely a definition, but motivated by observation.

Whenever we see a sequence of numbers coming closer and closer to a rational number $x$ (precisely: arbitrary close, and staying there), then this number $x$ is called the limit of the sequence. This again is just a definition.

Having these definitions in place, one then discovers, that there are sequences which have strange properties:

  • the sequence behaves like a convergent sequences of rational numbers in many aspects, e.g. it is making shorter and shorter steps over time, etc. (see Cauchy sequence)
  • there seems to be no rational number which this sequences comes closer and closer to.

So we have two options. We either accept that there are strange "convergent" sequences that do not converge against anything, OR we invent new numbers that fill these gaps. These numbers are the irrational numbers. E.g.

$$3+0.1+0.04+0.001+0.0005+0.00009+\cdots =: \pi.$$

So I recommend thinking about it the other way around. There are no strange rational sequences/sums that suddenly are no longer rational, but the irrational number was placed precisely there were it is in order to give this sequence a limit in the first place. The surprising observation therefore is that without this irrational number in place, there is nothing this sequence/series will eventually converge to. Maybe accepting that some specific infinite sums do not converge at all (while also not oscillating or diverging to $\pm\infty$) is more comprehendable.

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I would stay away from explanations involving infinite series, as that just adds another "mystical math thing" to deal with.

How about: start with the definition of a rational number, i.e. ratio of 2 integers, then run through the simple proof (by contradiction) that $\sqrt{2}$ is irrational?

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  • $\begingroup$ That makes sense. $\endgroup$ – Lance Pollard Jun 20 '18 at 15:28
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Cut a string so that its length is that of the diameter of the wheel. Wrap the string along the wheel's edge and make a mark at each endpoint. Then use it to measure the curve's length from the clockwise endpoint, going further in the clockwise direction a distance equal to the string's length, and make the next mark. And the next in the same way. And so on. After measuring $22$ times the length of the string, you've gone around the circle approximately seven times and you're close to your starting point, but not exactly at the same point. Keep going. Will you ever hit the same point twice? That $\pi$ is irrational just means that you won't.

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  • $\begingroup$ I like this answer because the explanation is actually suitable to a layperson. But, will they understand the significance of the mark lining up? $\endgroup$ – Douglas Held Jun 20 '18 at 19:12
  • $\begingroup$ @DouglasHeld : I think some will and some won't. $\endgroup$ – Michael Hardy Jun 20 '18 at 21:23
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Just because you have a sequence $a_n$ that gets closer and closer to $a$, that doesn't mean $a_n$ and $a$ have to be the same type. For instance, you can have a sequence of polygons that gets closer and closer to a circle, but the circle isn't a polygon. If you have a sequence of staircases where the size of each step gets smaller and smaller, they will get closer and close to a diagonal line, even though none of the staircases is a diagonal line. If you take a number and take half of it, then half of that, and so on, then every number you get will be positive, but you will get closer and closer to zero, which is not positive.

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Representations that use ... to denote an infinite sequence often trick the mind into thinking the infinite sequence will behave like a finite one. They don't.

The fraction series are a neat way to give you an idea of the value of an irrational. If cut the sequence somewhere and compute the result of that finite sum, you will get an approximation of the irrational number you are looking for. If you add the next term of the sequence, you will obtain a (usually) better approximation. This process can be repeated as many times as you like and you will get arbitrarily close to the value of the irrational. The limit of the sum to infinity is equal in the strictest sense to the irrational number. However, whether an infinite sum is equal to its limit is more a philosophical question than a mathematical one. This is often defined to be true for convenience.

In general the limit of a sequence isn't always defined because the sequence might not converge. Maybe this makes it easier to accept that sequences might converge towards things that have properties absent of the elements of the sequence.

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I would say:

An irrational number is a number that is impossible to measure exactly. Imagine you had a nut and some wrenches in all the standard sizes. This nut would never fit perfectly, no matter how many wrenches you had. For instance, what if it was too small for the 1/2", but too big for 1/4". But neither was it suitable for 3/4", and then you get into 5/8" etc. and 9/16" and even 11/64". No matter how fine of a wrench size you had, the nut would always be in between two sizes and never match exactly. Even if you had a crazy size like 2/3", such a wrench would also never fit.

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    $\begingroup$ An irrational number is a number that isn't equal to any rational number. Kinda begs the question, doesn't it? And if you're using a binary system, like you use in your paragraph, even rational numbers like 1/3 can't be measured exactly. $\endgroup$ – Acccumulation Jun 20 '18 at 21:07
  • $\begingroup$ Oh, I realize why all the answers are so "bad". Now I actually read the question, not just the title. My answer does not address the question. $\endgroup$ – Douglas Held Jun 22 '18 at 3:40

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