# How would one find other real numbers that aren't in the rational field of numbers?

For example, $\sqrt2$ isn't a rational number, since there is no rational number whose square equals two. And I see this example of a real number all the time and I'm just curious about how you can find or determine other numbers like so. Or better yet, how was it found that if $\mathbb Q$ is the set of all rational numbers, $\sqrt2\notin\mathbb Q$?

I appologize if the number theory tag isn't appropriate, I'm not really sure what category this question would fall under.

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You should write $\sqrt{2}\notin \mathbb{Q}$ insetad of $\sqrt{2} \neq \mathbb{Q}$. – Oliver Braun Jan 25 '13 at 17:19

I'm not sure if you're asking about finding a real number, or determining whether a given real number is rational or not. In any case, both problems are (in general) very hard.

Finding a real number

There are lots and lots of real numbers. How many? Well the set of all real numbers which have a finite description as a string in any given countable alphabet is countably infinite, but the set of all reals is uncountably infinite $-$ if we listed all the reals that we could possibly list then we wouldn't have even scratched the surface!

Determining whether or not a given real number is rational

No-one knows with certainty whether $e+\pi$ or $e\pi$ are rational, though we do know that if one is rational then the other isn't. In general, finding out if a real number is rational is very hard. There are quite a few methods that work in special cases, some more sophisticated than others.

An example of a ridiculous method that can be used to show that $\sqrt[n]{2}$ is irrational for $n>2$ is as follows. Suppose $\sqrt[n]{2} = \dfrac{p}{q}$. Then rearranging gives $2q^n=p^n$, i.e. $$q^n + q^n = p^n$$ but since $n>2$ this contradicts Fermat's last theorem. [See here.]

The standard proof of the irrationality of $\sqrt{2}$ is as follows. Suppose $\sqrt{2} = \frac{p}{q}$ with $p,q$ integers and at most one of $p$ and $q$ even. (This can be done if it's rational: just keep cancelling $2$s until one of them is odd.) Then $2q^2=p^2$, and so $2$ divides $p^2$ (and hence $p$); but then $2^2$ divides $2q^2$, and so another $2$ must divide $q^2$, so $2$ divides $q$ too. But this contradicts the assumption that one of $p$ and $q$ is odd.

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You might be interested in this earlier post, where the question being answered is "How do we prove that for any $a \in \mathbb{N},\;\sqrt{a}$ is an integer, or else irrational"? Some of the answers generalize even further.

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We take all cauchy sequences whose elements are in rationals. now there are some sequences whose limit is not in the rationals(We call this property as rationals are not complete.) and we add all the limits to a number system and we call this number system as reals.

I don't know your background and so i am unable to explain all the technicalities.

You can refer Analysis-1 by Terence Tao for more details. Or you can check Dedekind cuts if you can understand (Dedekind cuts is another way to create reals from rationals.)

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In the flavor of $\sqrt{2}$, any number which is not a perfect square (i.e. not the square of an integer), has an irrational square root. But how to prove this? One proves this as one proves that many numbers are irrational (as per your question), by assuming they are rational, and then showing that this leads to a contradiction.

The Ancient Greeks proved that $\sqrt{2}\notin \mathbb{Q}$ in the following way. Assume that $\sqrt{2}$ is rational. Then there exist $a,b\in \mathbb{Z}$ so that $$\frac{a}{b}=\sqrt{2}$$ where $a$ and $b$ are relatively prime, which implies that $$a^{2}=2b^{2}$$. Then clearly $a^2$ is divisible by $2$. This then implies that $2|a$ so that for some $c\in \mathbb{Z}$, $a=2c$ so that $$4c^{2}=2b^{2}$$ and $$b^{2}=2c^{2}$$ and $2|b^{2}$ so that $2|b$. Then $2|a$ and $2|b$ so that $a$ and $b$ are not relatively prime, contrary to assumption. So $\sqrt{2}\notin \mathbb{Q}$.

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Sorry I don't quite get what it means that $a$ and $b$ are relatively prime. I have a feeling that's some number theory, which I haven't had. And also, the $2|b^2$ is implying a cut correct? – TheHopefulActuary Jan 26 '13 at 20:02
If $a$ and $b$ are relatively prime, then their greatest common denominator is 1. For instance, 2 and 3 are relatively prime. So are 2 and 9. $2|b^{2}$ is notation which says that 2 divides $b^{2}$. So, for instance, if you let $b=4$, then 2 divides $4^{2}=16$ or $2|4^{2}$ – Jebruho Jan 27 '13 at 6:51

Perhaps something that will answer your " I'm just curious about how you can find or determine other numbers like so" So this will not explain how to find all of them, but I guess it motivates why $\sqrt{2}$ is irrational. So take an arbitrary polynomial $P$ with integer coefficients and let's find its roots.

Let's say $\deg P = 1$. Then $ax + b = 0$. We have the solutions are always rational.

Now what about $\deg P = 2$? It it doesn't factor over the rationals (i.e. $ax^2 + bx + c = a(x-r_1)(x-r_2)$ for some $r_1,r_2$ rationals) then clearly its roots are not rational.

However, let's take the field $\mathbb{Q}[x]_{P(x)}$. This is clearly a field extension of $\mathbb{Q}$ (well strictly speaking it isn't, but it is for our purpose). So this field is basically the set of all $m + nx$ for $m,n$ rational such that $ax^2 + bx + c$. It turns out by the Fundamental Theorem of Algebra that this set is a subset of $\mathbb{C}$. Letting $ax^2 + bx + c = x^2 - 2$ we get the familiar $\sqrt{2}$ (the reason why we did all this "unnecessary" work to define $\mathbb{Q}[x]_{P(x)}$ is to show that $x^2 - 2$ indeed has a root in some field which contains $\mathbb{Q}$).

Now take any other (irreducible) polynomial. By an identical process to above we can find many more irrational numbers. The numbers obtainable from this process are just the algebraic numbers. Of course, not all irrational numbers are algebraic because there are things such as $\pi$. Defining the transcendental numbers is much trickier and does not relate much to how $\sqrt{2}$ is defined.

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