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I'm trying to do some proof problems that includes irrational numbers. In order to do the proof, I need to do definitions first. Before this, I was doing proof about odd numbers (which can be defined as $2k + 1$, any integer $k$), even numbers ($2k$), and rational numbers ($a/b$, any integers $a$ and $b$, $b$ cannot be $0$).

So I know those definitions... But I'm not sure how to define irrationals?

It's kinda tricky for me, I would appreciate any help. Thanks!

lioness99a
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Niko H
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    A real number is irrational if and only if it is not rational. – YiFan Tey Feb 05 '20 at 05:20
  • But what are real numbers? I think the problem is that only after enough time studying mathematics one tends to get the hang of their construction. I mean (sequences)/(sequences to 0) feels not as canonical as rationals, even though you make a quotient construction for the latter too... – Jonas Linssen Feb 05 '20 at 06:42
  • @YiFan: this statement is useless as long as the reals have not been defined. –  Feb 05 '20 at 08:56
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    I would have given the same answer as YiFan. I'm not sure the OP has enough mathematical sophistication yet to really understand the construction of $\mathbb{R}$ out of $\mathbb{Q}$. What is instead often done in a first analysis class is to introduce the real numbers axiomatically - i.e. as a complete ordered field. (Of course, existence has to be shown but as a first introduction I believe this suffices.) – Qi Zhu Feb 05 '20 at 09:31

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This is done in an unexpected, and fairly technical way.

A real number is defined as a set of rational numbers (yes, a set). Such a set set must have two properties:

  • any element to the left of an element ($<$) is also an element, and

  • any element has an element to its right ($>$).

A first example is:

$$A:=\{q\in\mathbb Q:q<0\}.$$

Indeed, $r<q<0\implies r<0$, and $q<r=\dfrac{q}2<0$ and the two properties are always fulfilled. But this case is uninteresting, as it essentially redefines the real number $0$ (by notational abuse), also a rational, already know of us.

We have something more substantial with

$$B:=\{q\in\mathbb Q:q<0\lor q^2<2\}.$$

Now, bypassing the case $q<0$, we have $0\le r<q\land q^2<2\implies r^2<2$, and $q^2<r^2=\left(\dfrac{3x^2+2}{4x}\right)^2<2$. This defines a real number, which we will denote as $\sqrt 2$, and which corresponds to no rational. This is our first irrational number.

It is important to notice that these definitions only involve rational numbers and properties of rational numbers. This process is called a Dedekind cut.

Finally, a rational number is a real having an upper bound, and an irrational, having no upper bound.

The upper bound of a set is a number larger than all elements in the set and such that "there is no number in between". For the set $A$, $0$ is larger than all its elements, and no negative number is also larger than all elements in $A$. For the set $B$, you cannot find an upper bound.

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    I did not give the downvote, but I think this, while being the technically correct answer to the question "how do you define irrational numbers", is not going to help the OP when they are fundamentally confused about the concept of numbers being irrational. (At least, that's the impression I get; it is very much possible that my assumption about the OP's level of understanding is wrong.) – YiFan Tey Feb 05 '20 at 12:08
  • @YiFan: how can you build any proof if you define an irrational as "not a rational", which means that you cannot count on any property of the rationals. –  Feb 05 '20 at 13:17
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    Again, you are completely correct from a formal, logical point of view. But your answer is probably not very useful for OP if my guess as to their level of mathematical understanding is correct. It's a bit like (and I'm stretching the analogy here) trying to teach how to work with differentials to a freshman learning calculus for the first time by teaching him differential geometry in all its generality and rigour; there's nothing formally wrong with it, but it's still not pedagogically a good idea. – YiFan Tey Feb 05 '20 at 14:03
  • @YiFan: what would you have answered to the question "what is a fractional number ?". Answering "not an integer" leads nowhere. The OP is asking an operational definition, not a vague intuition. –  Feb 05 '20 at 14:06
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    I'm not sure why you think I would answer "not an integer" to that question; the usual context that one works in throughout high school is the set of reals (which is taken for granted without formal construction), so saying "not an integer" would lead to such a student thinking about irrational numbers as well. – YiFan Tey Feb 05 '20 at 14:18
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    OP did not post their "proof problems" so we can only guess, but I would expect that it is probably something along the lines of showing a certain number is irrational (such as the classic irrationality proof of $\sqrt2$). These are based on showing no $a,b\in\mathbb Z$ exist with $a/b=\sqrt2$, and that's a perfectly "operational" definition. Is it also a little bit unrigorous if we haven't defined $\mathbb R$? Perhaps, but that's really besides the point here, imho. I hope you can see my point here; at risk of turning this into a long discussion, this will be my last word on the subject. – YiFan Tey Feb 05 '20 at 14:18
  • @YiFan: what would you answer to "what is a fractional number" ? –  Feb 05 '20 at 14:48