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.
Best Answer
We need to prove that
Using contrapositive, we can instead prove
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.