Irrational + irrational = rational

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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?
 
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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
 
It's possible, certainly. The probability is 0 though.
 
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.
 
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.
 
When I say a random number, I mean one that can be inifinitely small to infinitely large (including number of digits) for both positive and negative numbers.
 
Jameson said:
When I say a random number, I mean one that can be inifinitely small to infinitely large (including number of digits) for both positive and negative numbers.

That doesn't tell me what you mean by random. Are all the numbers equally likely? What is the distribution?
 
Sorry. Yes, I am saying all numbers are equally likely... I don't know what you are asking by distribution though.
 
Jameson said:
Sorry. Yes, I am saying all numbers are equally likely... I don't know what you are asking by distribution though.

http://mathworld.wolfram.com/DistributionFunction.html

That is not possible. When you say that the numbers are equally likely then you are saying that you are using a uniform distribution. You can’t have a uniform distribution on an unbounded set