Irrational + irrational = rational

[tonebone10]

Can someone prove that there exists x and y which are elements of the reals such that x and y are irrational but x+y is rational? Certainly, there are an infinite number of examples (pi/4 + -pi/4 for example) to show this, but how would you prove the general case?

 

[Jameson]

$$\frac{\pi}{4} + \frac{3\pi}{4} = \pi$$

This is not a rational number. I don't believe you can add two irrational numbers and get a rational result. The only case I could think of would be something like $$\sqrt{5} - \sqrt{5} = 0$$.

I know you can multiply two irrational numbers to get a rational one, but like I said, I don't think addition can do this.


Jameson

 

[tongos]

ofcourse you can,
how about
(In6)/(In3) and (In1.5)/(In3) they add to give 2.

 

[Jameson]

Hmmm... good example tongos. Can you give an example though that doesn't have any division to express the irrational number?

 

[tongos]

how about 2 sqrts that add to equal one.

sqrt(x)+sqrt(y)=1
y=1-sqrtx+x if x is rational, and not a perfect square then y can't be rational.
sqrty is irrational

 

[Icebreaker]

Jameson,

What about $$(\pi -1)+1$$?

Although I don't know if $$(\pi -1)$$ can be considered as an irrational number, or a "number".

 

[Jameson]

Pi is irrational.

Tongos - I don't understand what you mean... please explain and give an example.

 

[Icebreaker]

I wrote it wrong. I meant $$(1-\pi)+\pi$$

 

[Jameson]

Yeah, I agree. If you look at my first post I gave an example like that. You're basically just cancelling out the irrational number... can you show me one where you add them and they don't result in zero?

 

[Icebreaker]

$$(1-\pi)+\pi=1$$

The question is whether $$(1-\pi)$$ can be considered as an irrational number.

 

[Jameson]

That is an irrational number and I agree it is a case that proves the inital topic. But what you're doing is saying $$a + C -a = C$$

This is where "a" is the irrational number and C is any rational constant. Can you show me an example where you don't simply subtract the irrational number to get a rational one?

 

[tongos]

if there is no pattern in the numbers after the decimal point in x,
then when you add it to y to obtain a rational number, y would also have to be irrational, or have a no pattern in its numbers.

 

[Jameson]

Very true.

 

[tongos]

jameson, my example about the sqrts shows it

 

[Jameson]

Yeah, I get it now. Thank you for explaining.

 

[Icebreaker]

What about this for a general case?

Assume that the difference between a rational number and an irrational number is an irrational number. The sum of any such two irrational numbers would be a rational number.

All we have to do now is prove the assumption, which I think is easier.

edit: I just saw tongos' proposition, which I think is better than mine.

 

[mathwonk]

this was a homerwork problem freshman year in about my first homework. My idea was to take any decimal with non repeating entries, hence irrational, and then add to it thed ecimal whose entries were 9-the entry in the firstd ecimal. that gives .9999999 ... =1 a rational number. finally i realixzed that this mjust emant thnat for any rational number r and irrational number, we have r-x and x both irrational, hence (r-x) + x = r is rational.

 

[Data]

I assume we can all agree with

$$\forall x, y \in \mathbb{R}, \ \exists z \in \mathbb{R} \ \mbox{s.t.} \ x + z = y$$

or, equivalently

$$x, y \in \mathbb{R} \Longrightarrow x-y \in \mathbb{R}$$

Then let $$\mathbb{I}$$ denote the set of irrationals. Clearly

$$a \in \mathbb{I}, \ b \in \mathbb{Q} \Longrightarrow a + b \in \mathbb{I}$$

since otherwise, we assume

$$a + b = d \in \mathbb{Q} \Longrightarrow \exists q, p, n, m \in \mathbb{Z} \ \mbox{s.t.} \ a + \frac{n}{m} = \frac{q}{p} \Longrightarrow a = \frac{qm - np}{pm} \in \mathbb{Q}$$

which is a contradiction.

Basically, the difference between a rational number and an irrational number is ALWAYS irrational.

 

[Icebreaker]

Although it does seem like circular logic:

"The sum of two irrational numbers is rational iff the difference between the sum and one of the irrational numbers is irrational."

 

[matt grime]

But how can it be any other way? If x+y=r with x, y, irrational and r rational then of course x=r-y, it isn't circulary logic, it's just bleedin' obvious.

 

[Icebreaker]

Yes, even though it is what I proposed and what Data has proven, I still don't know whether if it answered the op's question (the "general case").

 

[matt grime]

Of course it answers the general case, where do I specify anything on x and y other than they are irrationals whose sum is a rational?

 

[Jimmy Snyder]

non-repeating decimal

Take any irrational x and display it as a non-repeating decimal:

$$x = a_{0}.a_{1}a_{2}...$$

$$a_{0}$$ can be any positive integer. then let y be:

$$y = b_{0}.a_{1}a_{2}...$$

where $$b_{0}$$ is any positive integer for which $$b_{0} \ne a_{0}$$

Then x and -y are both irrational, x + (-y) is rational, and nothing in sight adds up to zero.

 

[matt grime]

apart from y-x and b_0-a_0, add them two things up...

 

[Icebreaker]

"The sum of two irrational numbers is rational iff the difference between the sum and one of the irrational numbers is irrational."

This can be re-worded as:

"The sum of the two irrational numbers is rational iff the sum of the two irrational numbers is rational."

Seems like redundancy to me.

 

[Data]

The sum of two irrational numbers is rational iff the difference between the sum and one of the irrational numbers is irrational.

Is wrong (and quite silly). You are essentially stating that the sum of two irrational numbers is never irrational.

$$\frac{\pi}{4}, \frac{\pi}{2} \in \mathbb{I} \; \mbox{and} \ \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \in \mathbb{I}$$

but

$$\frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4} \in \mathbb{I}$$

Thus the statement can clearly not be "re-worded as" (if you read this as "equivalent to")

The sum of the two irrational numbers is rational iff the sum of the two irrational numbers is rational.

which is true.

There is no useful, obvious "if and only if" statement here. The best we can do is:

$$\mbox{If} \ a \in \mathbb{I} \ \mbox{and} \ a+b \in \mathbb{Q} \ \mbox{then} \ b \in \mathbb{I}$$

and

$$\mbox{If} \ c \in \mathbb{Q} \ \mbox{then} \ \forall a \in \mathbb{I}, \ \exists b \in \mathbb{I} \ \mbox{s.t.} \ a+b=c$$

the second statement being completely obvious given the first, and the properties of the real numbers.

 

[Icebreaker]

Is wrong (and quite silly). You are essentially stating that the sum of two irrational numbers is never irrational.

Yes, I see the error. The statement works if we replace the "iff" with "if", although it still cannot be equivalent to the second.

 

[Zurtex]

The sum of any such two irrational numbers would be a rational number.

If you take any 2 random real numbers, a and b, the chance of their sum being rational is 0.

 

[Data]

The statement works if we replace the "iff" with "if"

No it doesn't. The counterexample in my last post still works. Replacing "iff" with "only if" works though.

 

[Icebreaker]

No it doesn't. The counterexample in my last post still works. Replacing "iff" with "only if" works though.

Yes, you're right. I tend to try and avoid math symbols in these statements, though; which is why I try to formulate phrases to explain the ideas, and they turn out, unfortunately, to be ambiguous or errorous.

If you take any 2 random real numbers, a and b, the chance of their sum being rational is 0.

Yes... if you take two RANDOM real numbers. But if we subtract a rational with an irrational, the difference will be irrational. Take these two irrational numbers, add them, and the sum will be rational.