Irrational + irrational = rational (2024)

[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

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

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".

Pi is irrational.

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

[tex](1-\pi)+\pi=1[/tex]

The question is whether [tex](1-\pi)[/tex] can be considered as an irrational number.

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?

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.

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.

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."

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.

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.

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.

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.

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.

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.

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.

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?

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

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.