# Proving Irrationals Are Dense in the Reals

someperson05

## Homework Statement

Show that the irrational numbers are dense in R. [Hint: Use the fact that sqrt(2) is irrational.]

## Homework Equations

Book already proved that the rationals are dense in R.

## The Attempt at a Solution

My intuition is that I'm supposed to make some sort of argument that since the rationals are dense in R, then just by transposing the rationals by sqrt(2), I'm showing that irrationals are also dense. (Using that a rational plus an irrational is an irrational.)
I just can't figure out for the life of me how to start this argument.

I could just say
x < r < y ->
x + sqrt(2) < r + sqrt(2) < y + sqrt(2) ->
a = x + sqrt(2), b = y + sqrt(2) ->
a < r + sqrt(2) < b

But that feels horribly unprooflike...and just quite horrible. I want something better than that. I also feel like this should be horribly easy, but I just can't seem to get anything worthwhile.

Give me as little information as you can please. If anything, I just need a point in the right direction, just a suggestion of how I should be venturing to prove this.

Last edited:

Staff Emeritus
Gold Member
Well, it might help to write things out in full. Shorthand is great when it's useful, but when we're confused, it often hides the details that make everything clear.

A good starting point is an explicit, mathematical statement of what it means for the irrationals to be dense in R. That way, we can clearly see what our target is.

Another good starting point is an explicitly stated theorem that would allow you to conclude a set is dense that we could try applying. But your current proof idea seems quite direct, so let's chase after that first before trying other starting points.

someperson05
I think I've got something.

Some previous results that I'm using is that a rational plus an irrational is irrational, and that a rational multiplied by an irrational is also irrational.

So since I knew that for any r that is an element of Q, and x, y that are elements of R, then
x < r < y. Or that the rational numbers are dense in R.

So consider an irrational number v, such that
v = r + s(sqrt(2))

Where s is an arbitrary rational number.
If we let s(sqrt(2)) < y - r, so that
s < (y - r)/sqrt(2)

Then it follows that
x < r + s(sqrt(2)) < y

And for any interval of real numbers, an irrational can be constructed that exists between them. So we can conclude that the irrationals are dense in R.

Does this seem like a valid proof? I like it, but I want to know if it is formal enough or not. Any critiques are more than welcome.

Where s is an arbitrary rational number.
If we let s(sqrt(2)) < y - r, so that
s < (y - r)/sqrt(2)

Then it follows that
x < r + s(sqrt(2)) < y

Not strict enough here. You're letting s be negative, which would contradict x < r + s(sqrt(2))

Staff Emeritus
Gold Member
So since I knew that for any r that is an element of Q, and x, y that are elements of R, then
x < r < y. Or that the rational numbers are dense in R.

You said:
for any r that is an element of Q, stuff
So, I'll choose r=0. If what you said is true, then stuff must be true for this choice of r. So what is stuff? It is
for any x,y that are elements of R, more stuff
I'll choose x=2 and y=1. If what you said is true, then more stuff must be true for these choices of x, y, and r. So what is more stuff?
2<0<1​
Alas.

Getting quantifiers wrong is a fairly common problem. In this case, you made two mistakes:
1. You got the ordering wrong: x and y were supposed to be chosen first.
2. You got one type wrong: the quantifier on r is existential, not universal. i.e. "there exists" not "for all". For each choice of x and y, density merely asserts there is at least one value of r that works -- not that every value of r works.

Now, you got the English prose right:
And for any interval of real numbers, an irrational can be constructed that exists between them.
but your actual proof was similarly backwards. What you actually proved was that, for any irrational number, you can construct an interval around it.

Getting it backwards is also a common problem; I suspect it's from doing basic algebra problems; when solving problems, you pretty much always "work backwards" -- you do work that proves "If this equation has a solution, then it is ______". People often forget to go on to prove that ____ actually is the solution.

Anyays, for clarity, the statement you want to prove is:
• For any pair of real numbers x,y:
• If x < y then:
• There exists an irrational number r such that:
• x < r < y

someperson05
Thank you Raskolnikov for pointing out my oversight.

Also a big thanks to you Hurkyl, you have provided some clarity.

So I hope this proof proceeds in the correct direction. I have just modified my backwards one.

For any pair of real numbers, x and y,
If x < y,
Then there is an irrational number r such that x < r < y.
(I'm hoping my statement of the problem is correct.)

I tried to proceed by using a proof by cases, and modifying my use of the variable s.

Assume that x is rational.

We know x < y
Now let a variable s be rational and positive.
x + s(sqrt(2)) is greater than x.
Further, since x is rational, x + s(sqrt(2)) is irrational.
Now if we let s(sqrt(2)) < y - x, then x + s(sqrt(2)) < y.
It follows that x < x + s(sqrt(2)) < y.
So we have shown between the variables x and y, there exists an irrational number.
(At least I think it shows that. I may be getting it backwards again. Instead of constructing the irrational from thin air like I did previously, I am trying to construct it from x, to show that it is in between both x and y.)

(Was my error before that I proceeded from the irrational to showing that x and y was around it, instead of beginning with x and y and showing that an irrational existed between them?)

Assume x is irrational.
Let a variable s be rational, positive and less than y - x. ( 0 < s < y - x )
x is less than x + s, since s is positive. ( x < x + s )
Since x is irrational and s rational, x + s is irrational.
Since s < y - x, x + s < y.
So x < x + s < y, and so it has been shown that an irrational exists between x and y.
(It doesn't seem like I need to assume anything about y, whether rational or irrational since I don't actually manipulate y in any manner.)

If the above proof is correct, I just had a thought that may simplify matters.
Could I have chosen to let s = y - x / 2? Or any arbitrary number on the bottom for that matter? Since the purpose is to just show that there exists an irrational between x and y, I just need to give one example.