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There must be an error in my "proof" since it is evident that the sum of two irrational numbers may be rational, but I am struggling to spot it. A hint would be appreciated.

The "proof" is by contradiction:

Assume that the sum of two irrational numbers a and b is rational. Then we can write

$$ a + b = \frac{x}{y} $$

$$ \implies a + b + a - a = \frac{x}{y} $$

$$ \implies 2a + (b - a) = \frac{x}{y} $$

$$ \implies 2a = \frac{x}{y} + (-1)(b + (-1)(a)) $$

-> from our assumption that the sum of two irrational numbers is rational, it follows that $(b + (-1)(a))$ is rational

-> therefore, the right side is rational, being the sum of two rational numbers

-> but the left side, $2a$, is irrational, because the product of a rational and irrational number is irrational

-> this is a contradiction; since assuming that the sum of two irrational numbers is rational leads to a contradiction, the sum of two irrational numbers must be irrational.

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    $\begingroup$ "from our assumption that the sum of two irrational numbers is rational". So you've gone from assuming that there are two irrationals with an irrational sum, to assuming that every two irrationals have a rational sum?! $\endgroup$ – Lord Shark the Unknown Nov 30 '19 at 5:59
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    $\begingroup$ It is false from the begining...$ e+(-e)=0$... $\endgroup$ – dmtri Nov 30 '19 at 6:00
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    $\begingroup$ Welcome to Math Stack Exchange. Write your questions clearly. Please use MathJax . See math.meta.stackexchange.com/questions/5020/… $\endgroup$ – nmasanta Nov 30 '19 at 6:09
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    $\begingroup$ This Q can be a good test for students. For millenia, back to & including Euclid, paradoxes have been presented, with the challenge being to find the hidden assumption or logical error. $\endgroup$ – DanielWainfleet Nov 30 '19 at 6:27
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    $\begingroup$ What you have proved, tough, is that if a+b and a-b are rational, then both a and b are rational. $\endgroup$ – Vincent Fourmond Dec 1 '19 at 14:49
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To say that it is not true that all swans are white does not mean that all swans are non-white; it only means that at least one swan is non-white.

Similarly, to say that it is not true that every sum of two irrational numbers is irrational does not mean that every sum of two irrational numbers is rational; it only means that at least one sum of two irrational numbers is rational.

You start by assuming, not that the sum of (every) two irrational numbers is rational, but rather that the sum of two irrational numbers $a$ and $b$ is rational, i.e. that there is one instance of two irrational numbers whose sum is rational.

That assumption is true. For example: If $a=\pi$ and $b=4-\pi,$ then the sum of the two irrational numbers $a$ and $b$ is the rational number $4.$ And the sum of the two irrational numbers $a$ and $-b$ is the irrational number $2\pi-4.$ The fact that the sum of two irrational numbers $a$ and $b$ is rational does not mean that the sum of the two irrational numbers $a$ and $-b$ is rational, nor that any other sum of two irrational numbers is rational.

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    $\begingroup$ "nor that any other sum of two irrational numbers is rational" -- Consider the sum of $2a$ and $2b$ ;) $\endgroup$ – Noiralef Dec 2 '19 at 8:12
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    $\begingroup$ An even simpler example is π and -π. (π-π) = 0 is rational, (π+π) = 2π is not. $\endgroup$ – RBarryYoung Dec 2 '19 at 15:48
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    $\begingroup$ @RBarryYoung : A simpler example but and inferior one. It is so simple that the reader might leap to the conclusion that the sum of two irrational numbers can be rational only in such a simple case. $\endgroup$ – Michael Hardy Dec 3 '19 at 16:09
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You have assumed $a+b$ and $b-a$ are rational and arrived at a contradiction. Therefore, the strongest conclusion your proof can make is at least one of $a+b$ and $b-a$ must be irrational.

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The initial assumption is, "Assume that the sum of two irrational numbers $a$ and $b$ is rational"; later on you say, "from our assumption that the sum of two irrational numbers is rational...", but these are not the same statement. You've cunningly morphed from an existential statement to a universal statement. That is: from an assumption that there are at least two numbers for which it is true (which is true), to an assumption that it's true for any such numbers (which is false).

A specific counterexample would be $a = -\sqrt 2$, $b = \sqrt 2$. In this case $a + b = 0$ is indeed rational, but $(b + (-1)(a)) = 2 \sqrt 2$ is not.

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    $\begingroup$ I read it not as assuming a universal "(a + b) is rational for all irrationals a and b" (which would be absurd), but a more specific "since (a + b) is rational, it follows that (b - a) is rational". I.e., that if there exist special pairs of irrationals that have a rational sum, any such pair also has a rational difference. But that's not only not necessarily true, it's easily provable that it can't be true for any such pair: (b - a) is ((a + b) - 2a). 2a is a rational times an irrational, which is irrational. So ((a + b) - 2a) is a rational plus an irrational, which is irrational. $\endgroup$ – dgould Dec 3 '19 at 23:52
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Sum of two irrational numbers can be rational or irrational. In your argument you are assuming the sum of any two irrational numbers is irrational and arriving at a contradiction. In you 'proof' there is no reason why $b+(-1)a$ is rational.

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There is another minor mistake in your proof.

The use of the $\implies$ symbol.

Observe that if one can write: $$a+b=\frac{x}{y}$$ $$\implies a+b+(a−a)=\frac{x}{y}$$ $$\implies a+b=\frac{x}{y}$$

independent of the truth of falsity of the premises if one can go back what one have to say is:$$a+b=\frac{x}{y}$$ $$\equiv a+b+a−a=\frac{x}{y}$$ because $P \equiv Q$ means $P\implies Q$ and $Q\implies P$.

The $\equiv$ operator emphasizes that one is transforming an equation to an equivalent form.

A more clear way to say it is:

Given any irrational numbers $a$ and $b$ there exist a rational number $\frac{m}{n}$ such $a+b=\frac{m}{n}$,

Proof by contradiction:

suppose that $a+b=\frac{m}{n}$

$a+b=\frac{m}{n}$

$\equiv$ ( because $x+(-x)=0$ and $x+0=0$ )

$a+b+(a−a)=\frac{m}{n}$

$\vdots$

This proof style may seem to detailed, but is more clear for the reader and the writer which helps to notice mistakes.

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You must have very clear the question that motivated the proof, i.e. which theorem you want to prove.

As answered before by others, you must pay attention to the quantification of what you want to prove.

What I want to add, is that you must understand if you really need a proof by contradiction.

You could have tried to build a counterexample by exploring the case of adding two irrational numbers could resulting in a rational number, something like exploring if you can compute the additive complement of the fractional part of a positive irrational, also an irrational. That number, which also is irrational, can be computed by the function $f(x) = [1-(x-\lfloor x\rfloor)]$, thus $x+f(x)=\lceil x\rceil \in \mathbb{N}$.

That is sufficient to prove that the sum of two irrationals can produce a rational number which also is a natural.

If your question were to know the set of all irrational numbers which produce a rational sum, you could keep working to search a function $g(x,m,n)=\cdots$ that given any $m,n\in \mathbb{N}$, produce an irrational which produce a rational number added to $x$, which could lead to $s(x)=\{g(x,m,n):m,n\in\mathbb{N}\}$.

With the above function you can even build the set of all pairs of an irrational and the corresponding set of irrationals, $(x,s(x))$. That set include the set of all the possible counterexamples, based on it, you can also build the set of all the pairs of irrationals whose sum is an irrational number, $t(x)=\{(x,\mathbb{R}\smallsetminus y):(x,y)\in s(x)\}$.

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