1
$\begingroup$

I am working on this proof and this is what I got so far, can someone help me verify if what I have done is right?

For all real numbers $x$ and $y$, if $x+y$ is rational and $x-y$ is irrational then $x$ is irrational and $y$ is irrational.

Proving the contrapositive and since contrapositive is logically equivalent to the original statement, we can conclude the original statement.

Contrapositive: For all real numbers $x$ and $y$, if $x$ is rational or $y$ is rational, then $x+y$ is irrational or $x-y$ is rational.

Proof: Suppose $x$ and $y$ are real numbers such that $x$ is rational. Then let $y = x$. We want to prove that $x+y$ is irrational or $x-y$ is rational. Since $x$ is rational, we know that $x = a/b$ where $a$ and $b$ are integers and $b \not= 0$. and $y = -x$ then $x-y = x-x = 0$, since $0$ is integer, and it is rational, that implies that $x-y$ is rational. End of Proof.

Now, can someone help me prove the original statement? Without trying with contrapositive?

$\endgroup$
2
  • $\begingroup$ Your proof is technically incorrect because you say "suppose $x$ and $y$ are real numbers" which implies that $x$ and $y$ are both arbitrary, but then you say "let $y = -x$" which implies that $y$ is not arbitrary. You also then wrote that $x - y = x-x$ but this is wrong because $y = -x$ implies $x - y = x-(-x) = 2x$ which is only $0$ if $x = 0$. $\endgroup$
    – john
    Jul 17, 2015 at 1:51
  • $\begingroup$ Thanks for pointing out that typo. $\endgroup$
    – Samim
    Jul 17, 2015 at 2:23

1 Answer 1

7
$\begingroup$

You can proceed directly as follows: $2x = (x+y) + (x-y)$ which must be irrational as it is the sum of a rational and an irrational. So $x$ is irrational. Similarly $2y = (x+y) - (x-y)$ is irrational.

$\endgroup$
12
  • $\begingroup$ Thanks John, but I don't understand. Where is the 2x coming from? $\endgroup$
    – Samim
    Jul 17, 2015 at 1:08
  • $\begingroup$ The original statement is: For all real numbers x and y, if x+y is rational and x-y is irrational then x is irrational and y is irrational. $\endgroup$
    – Samim
    Jul 17, 2015 at 1:09
  • $\begingroup$ In my answer I'm showing that given the original statement, I can conclude that $2x$ must be irrational. And $2x$ is irrational if and only if $x$ is irrational. $\endgroup$
    – john
    Jul 17, 2015 at 1:13
  • $\begingroup$ Based on the original statement, you only know that x and y are real numbers, such that x+y is rational and x-y is irrational. You do not know 2x=(x+y)+(x-y) - am I missing anything? $\endgroup$
    – Samim
    Jul 17, 2015 at 1:16
  • 2
    $\begingroup$ @Samim: but you are clearly allowed to use addition/subtraction, otherwise you couldn't write expressions such as $x+y$ and $x-y$. $\endgroup$
    – john
    Jul 17, 2015 at 1:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .