For two single irrational numbers "a" and "b" if (a+b) or (a-b) is rational , will it always be "0"?

Question

As per me the above statement should be true since if the sum of two irrational numbers is 0 , it implies that two irrational numbers must be like +k and -k where k is an irrational number . Also when diffrence of two irrational numbers is rational it is possible only when both the irrational numbers are same . From both the above cases we can conclude that if sum of two irrational numbers is rational or difference of two irrational numbers is rational then it must be 0 . Please correct me if I am wrong . By "a" and "b" I mean single or individual irrational numbers for example only "pi" and not "pi +/- something" .

Answer 1

Don't overthink it.

If $x$ irrational then $q - x$ is irrational for every rational $q$.

So $x + (q-x) = q$ is the rational sum of two irrational numbers. But $q$ can be any rational number and doesn't have to be $0$.

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This is one of those results that when you analyze it you realize it can't actually mean anything.

We think. $rational + rational = rational$

And $rational + irrational = irrational$.

But what of $irrational + irrational$ can that be rational.

Well, $\sqrt 2 + (-\sqrt 2) = 0$ and $0$ is rational so it can be rational.

So we think... and we say... Yeah, but $\sqrt 2$ and $-\sqrt 2$ are related to each other, they are negatives of each other. I bet if you have two independent irrationals the must add to a irrational.

So someone else says well if $x +y = k$ and $k$ is rational, then if $x$ is irrational then $y= k - x$ is irrational and $x + y = x+(k-x) = k$. so that is two irrationals that add to a rational.

So we say... but those are dependent to each other; I bet if you have two independent irrationals the must add to a irrational.

So the other person asks. How exactly to you define if two irrationals are "dependent"?

And we say... well if $x$ is irrational and $y = \pm x + k$ for some rational $k$ they are dependent. So I claim that if $x,y$ are independent than $x + y$ is not rational.

And then other person says.... so in other words, you are claiming that there is no rational number $k$ so that $y =\pm x + k$ then $y \mp x \ne k$ for any rational $k$ but if there is such a rational $k$ so that $y =\pm x + k$ then $y\mp x$ might be rational; is that what you are claiming?

And we say, yes, now how can I prove it?

And the other person says. Um... you just said $x \mp y =k; k$ rational if and only if $x = \pm y +k$ for some rational $k$. Do you really think that is meaningful?

And we say.... Ulp, I guess not.

Answer 2

If $a+b$ and $a-b$ are both rational, then so is $a$ since it is the average of those two quantities, and the average of two rational numbers is also rational. Thus also $b=(a+b)-a$ is rational.

Answer 3

Your statement in the body is not correct. For example, both $\sqrt 2$ and $\sqrt 2 +1$ are irrational. Their difference is $1$, which is rational. If both the sum and difference of two numbers are rational, the two numbers must be rational, but that is not what your question says.