Irrational + irrational = rational

In summary, the conversation discusses the existence of two irrational numbers, x and y, whose sum is a rational number. Several examples are given, including pi/4 and 3pi/4, and sqrt(x) and sqrt(y) where x and y are both irrational and their sum is 1. The conversation also explores the possibility of adding two irrational numbers and getting a rational result, but it is concluded that this is not possible unless one of the numbers cancels out. Finally, it is stated that the difference between a rational number and an irrational number is always irrational, and this is used to prove that the sum of two irrational numbers is rational if and only if the difference between the sum and one of the irrational numbers
  • #1
tonebone10
4
0
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?
 
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  • #2
[tex]\frac{\pi}{4} + \frac{3\pi}{4} = \pi[/tex]

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 [tex]\sqrt{5} - \sqrt{5} = 0[/tex].

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
 
  • #3
ofcourse you can,
how about
(In6)/(In3) and (In1.5)/(In3) they add to give 2.
 
  • #4
Hmmm... good example tongos. Can you give an example though that doesn't have any division to express the irrational number?
 
  • #5
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
 
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  • #6
Jameson,

What about [tex](\pi -1)+1[/tex]?

Although I don't know if [tex](\pi -1)[/tex] can be considered as an irrational number, or a "number".
 
  • #7
Pi is irrational.

Tongos - I don't understand what you mean... please explain and give an example.
 
  • #8
I wrote it wrong. I meant [tex](1-\pi)+\pi[/tex]
 
  • #9
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?
 
  • #10
[tex](1-\pi)+\pi=1[/tex]

The question is whether [tex](1-\pi)[/tex] can be considered as an irrational number.
 
  • #11
That is an irrational number and I agree it is a case that proves the inital topic. But what you're doing is saying [tex] a + C -a = C[/tex]

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?
 
  • #12
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.
 
  • #13
Very true.
 
  • #14
jameson, my example about the sqrts shows it
 
  • #15
Yeah, I get it now. Thank you for explaining.
 
  • #16
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.
 
  • #17
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.
 
  • #18
I assume we can all agree with

[tex] \forall x, y \in \mathbb{R}, \ \exists z \in \mathbb{R} \ \mbox{s.t.} \ x + z = y [/tex]

or, equivalently

[tex] x, y \in \mathbb{R} \Longrightarrow x-y \in \mathbb{R}[/tex]

Then let [tex] \mathbb{I}[/tex] denote the set of irrationals. Clearly

[tex] a \in \mathbb{I}, \ b \in \mathbb{Q} \Longrightarrow a + b \in \mathbb{I} [/tex]

since otherwise, we assume

[tex] 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}[/tex]

which is a contradiction.

Basically, the difference between a rational number and an irrational number is ALWAYS irrational.
 
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  • #19
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."
 
  • #20
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.
 
  • #21
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").
 
  • #22
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?
 
  • #23
non-repeating decimal

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

[tex]x = a_{0}.a_{1}a_{2}...[/tex]

[tex]a_{0}[/tex] can be any positive integer. then let y be:

[tex]y = b_{0}.a_{1}a_{2}...[/tex]

where [tex]b_{0}[/tex] is any positive integer for which [tex]b_{0} \ne a_{0}[/tex]

Then x and -y are both irrational, x + (-y) is rational, and nothing in sight adds up to zero.
 
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  • #24
apart from y-x and b_0-a_0, add them two things up...
 
  • #25
Icebreaker said:
"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.
 
  • #26
Icebreaker said:
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.

[tex] \frac{\pi}{4}, \frac{\pi}{2} \in \mathbb{I} \; \mbox{and} \ \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \in \mathbb{I}[/tex]

but

[tex] \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4} \in \mathbb{I}[/tex]

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:

[tex]\mbox{If} \ a \in \mathbb{I} \ \mbox{and} \ a+b \in \mathbb{Q} \ \mbox{then} \ b \in \mathbb{I}[/tex]

and

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

the second statement being completely obvious given the first, and the properties of the real numbers.
 
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  • #27
Data said:
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.
 
  • #28
Icebreaker said:
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.
 
  • #29
Icebreaker said:
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.
 
  • #30
Data said:
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.

Zurtex said:
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.
 
  • #31
Zurtex said:
If you take any 2 random real numbers, a and b, the chance of their sum being rational is 0.


What do you mean by random?
 
  • #32
It is possible to have the sum of two random numbers equal a rational one, although the chances of it happening are extremely minute.

If you are choosing two random numbers, then it is possible for the random number to be any number. This means the two random numbers could be [tex]\frac{\ln6}{\ln3}[/tex] and [tex]\frac{\ln1.5}{\ln3}[/tex].

As tongos pointed out, these two numbers add together to get the rational number 2. I understand that the chances of a random number being infinitely equivalent to the examples I gave are so small it wouldn't happen, but it's possible.


Jameson
 
  • #33
It's possible, certainly. The probability is 0 though.
 
  • #34
Jameson said:
It is possible to have the sum of two random numbers equal a rational one, although the chances of it happening are extremely minute.

That depends on what you (they/whomever) mean by random. When you say random I need to know what probability measure you are talking about. The real numbers have infinite support so you can't have a uniform distribution.
 
  • #35
I disagree. Probability is [tex]\frac{desired outcome}{all possible outcomes}[/tex]. You are looking for 2 outcomes in the instance, out of a possible outcomes of infinity.

The [tex]\lim_{x\rightarrow\infty}\frac{2}{x} = 0[/tex], yes I agree, but using the definition of probability, there is a chance it could happen in the possible number of outcomes, therefore it has a probability that is greater than 0.
 
<h2>1. What is the definition of irrational numbers?</h2><p>Irrational numbers are numbers that cannot be expressed as a ratio of two integers. They are non-terminating and non-repeating decimal numbers.</p><h2>2. How do irrational numbers behave when added to other irrational numbers?</h2><p>When two irrational numbers are added, the result can be either rational or irrational. It depends on the specific numbers being added.</p><h2>3. Is it possible for an irrational number to be added to a rational number and result in a rational number?</h2><p>Yes, it is possible for an irrational number to be added to a rational number and result in a rational number. For example, 1.5 (a rational number) can be added to √2 (an irrational number) to get 2.5 (a rational number).</p><h2>4. Can two irrational numbers be multiplied together to get a rational number?</h2><p>Yes, it is possible for two irrational numbers to be multiplied together and result in a rational number. For example, √2 and √8 are both irrational numbers, but when multiplied together, the result is 4 (a rational number).</p><h2>5. What is an example of an equation where irrational + irrational = rational?</h2><p>An example of an equation where irrational + irrational = rational is √2 + (-√2) = 0. This is because the irrational numbers, √2 and -√2, cancel each other out to equal a rational number, 0.</p>

1. What is the definition of irrational numbers?

Irrational numbers are numbers that cannot be expressed as a ratio of two integers. They are non-terminating and non-repeating decimal numbers.

2. How do irrational numbers behave when added to other irrational numbers?

When two irrational numbers are added, the result can be either rational or irrational. It depends on the specific numbers being added.

3. Is it possible for an irrational number to be added to a rational number and result in a rational number?

Yes, it is possible for an irrational number to be added to a rational number and result in a rational number. For example, 1.5 (a rational number) can be added to √2 (an irrational number) to get 2.5 (a rational number).

4. Can two irrational numbers be multiplied together to get a rational number?

Yes, it is possible for two irrational numbers to be multiplied together and result in a rational number. For example, √2 and √8 are both irrational numbers, but when multiplied together, the result is 4 (a rational number).

5. What is an example of an equation where irrational + irrational = rational?

An example of an equation where irrational + irrational = rational is √2 + (-√2) = 0. This is because the irrational numbers, √2 and -√2, cancel each other out to equal a rational number, 0.

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