Question: Prove that if the sum of two numbers is irrational then at least one of the numbers is irrational. Is your proof direct, by contradiction, or by contrapositive?

State the converse. Prove or disprove the converse.

For the first question I'm going to use a direct proof. Please let me know what method is easiest.

Proof:Suppose that the sum of two numbers is irrational that is $z = x + y$ where $z$ is irrational.

Case 1: Let x be the rational number and y the irrational number. That is $x = \frac ab$ where $a$ and $b$ are integers and $b$ can't be equal to $0$. Then $z = x + y = (\frac ab) + y = \frac{a + by}b$. Then $(a + by)$ is not an integer since $y$ is irrational and therefore $z$ is irrational.

Case 2: Let $x$ and $y$ be irrational. Thats is let $x$ does not equal to $(\frac ab)$ and $y$ does not equal to $(\frac cd)$ where $a,b,c,d$ are integers and $b,d$ cannot equal to $0$. Then $z = x + y$ doesn't equal $(\frac ab) + (\frac cd)$ doesn't equal to $\frac{ad+bc}{bd}$. Since $ad + bc$ is an integer and $bc$ is an integer then $\frac{ad + bc}{bd}$ is rational but $z = x + y$ does not equal to $\frac{ad + bc}{bd}$ therefore $z$ is irrational.

I'm not sure if this is the right way to prove this problem.

The converse is: If at least one of two numbers is irrational then their sum is irrational. Is this right? I think the converse is true but my proof is pretty similar as the above one.

I just consider two cases

Case 1. Both numbers are irrational

Case 2. One of the numbers is irrational.

Please let me know if my proof is correct or if there's a better way to prove it. Any hints or suggestions are welcome.

  • 4
    $\begingroup$ The easiest way to do the first part is to use a proof by contradiction: assume the two numbers are both rational, and show that this leads to a contradiction. You have stated the converse correctly, but your conjecture about this (and your reasoning in Case 2) isn't right. The sum of two irrational numbers can be a rational. $\endgroup$ – user84413 Sep 4 '13 at 0:51
  • $\begingroup$ Can I prove the converse by contradiction too where I assume that the sum is rational? Or is the converse false? $\endgroup$ – Candy Pelagio Sep 4 '13 at 1:34
  • 1
    $\begingroup$ The converse is false. Contraposition can be used to avoid contradiction while using the same steps. I.e., show that $a,b\in\mathbb Q\implies a+b\in\mathbb Q$. $\endgroup$ – Jonas Meyer Sep 4 '13 at 2:28
  • $\begingroup$ okay thank you. But I'm a little bit confused, if the converse is false why is the first statement true? Is it because we are just trying to show that for the sum of two numbers to be irrational then at least one of the two numbers is irrational without considering the summation of two irrational numbers? And for the converse we have to consider both cases which makes the converse false? $\endgroup$ – Candy Pelagio Sep 4 '13 at 3:07
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    $\begingroup$ @Candy: The first statement is true because the sum of 2 rational numbers is rational. The converse is false because a sum of 2 irrational numbers can be rational. $\endgroup$ – Jonas Meyer Sep 4 '13 at 3:13

We need to prove that

If $z=x+y$ is irrational, then either $x$ or $y$ is irrational.

Using contrapositive, we can instead prove

If $x$ and $y$ are rational, then $z=x+y$ is rational

Which you have proved in case (1).

The converse is not true since sum of two irrational numbers may not be irrational: for example, if $x$ and $y$ are irrational in such a way that $x=-y$, $x+y=0$ is rational.


From two rational numbers $\frac{n_1}{d_1}$, $\frac{n_2}{d_2}$, you can construct the sum $\frac{n_1*d_2 + n_2*d_1}{d_1*d_2}$ which is rational. You've just proven that the $\mathbb{Q}$ is closed under addition.

The opposite of "at least one number" is "neither number". Assuming the sum is irrational, either at least one number is irrational or neither number is irrational.

If neither number is irrational, then they are both rational and their sum is rational, which is a contradiction. Therefore, the former must be true that at least one number is irrational.


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