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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" .

Sameer Nilkhan
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    Your title talks about the sum and difference being rational, but the body talks about the sum or difference being rational. These are very different. The body is longer, so seems to be what you are asking about. Please clarify. – Ross Millikan Jan 18 '20 at 06:27
  • $\implies$ and $\Leftarrow$ are not the same thing. If $x + k =0$ then $k=-y$ is $k$ irrational. So if the some of an irrational and a number is $0$ the other number is irrational. That's $\Leftarrow$. That that WASN'T the question. The question was if $x$ is irration and $y$ is irrational but $x + y=q$ is rational is $q$ equal to $0$. That's $\implies$ and the answer is no. Take $x = \pi$ and $y = 2-\pi$. Both are irrational and $x + y = 2$ which is rational but not $0$. – fleablood Jan 18 '20 at 06:52
  • "if the sum of two irrational numbers is 0" That was not the hypothesis so it is not relevent. The hypothese was the sum of the two irrational number is rational. There are other rational numbers than just $0$...."Also when diffrence of two irrational numbers is rational it is possible only when both the irrational numbers are same" Nonsense! If $x - y = 2$, then does that mean $x$ and $y$ must both be rational? What if $x = \pi$? Then $y = \pi-2$. $\pi$ is irrational. Does that mean $\pi-2$ must be rational? – fleablood Jan 18 '20 at 07:00
  • @fleablood symbolically by "a" and "b" I mean single irrational number , that is , for example , only "pi" and not "pi - 2" . I do have sense just forgot to clarify what I mean – Sameer Nilkhan Jan 18 '20 at 07:24
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    $\pi -2$ is a single irrational number. How do you distinguish between $3.1415926535897932384626433832795....$ being a "single irrational number" but $\pi -2 = 1.1415926535897932384626433832795...$ is not? Are you saying $1.1415926535897932384626433832795...$ doesn't count? Isn't really irrational? Doesn't exist on its own? What? In any event $3.1415926535897932384626433832795...$ is irrational. And $1.1415926535897932384626433832795...$ is irrational. And $3.1415926535897932384626433832795...-1.1415926535897932384626433832795... = 2$ – fleablood Jan 18 '20 at 07:29
  • Note: $x + y = k\iff x = k-y$. So distinguishing between $x$ as an irrational number and $k-y$ as "not a single irrational number" doesn't really make any sense. – fleablood Jan 18 '20 at 07:31
  • OK you win , i was wrong thanks for your valuable time and guidance. – Sameer Nilkhan Jan 18 '20 at 07:32

3 Answers3

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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.

pre-kidney
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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.

Ross Millikan
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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 $0$.

=====

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.

fleablood
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