5
$\begingroup$

This is a homework assignment, please tell me if my proof is correct!

Prove that for every rational number z and every irrational number x, there exists a unique irrational number such that x + y = z.

Assume a and b are integers with GCD (a,b) =1, and that c is an irrational number. There exists a number d = (a/b) + c.

Assume that d is rational. Then d - (a/b) would be rational, what is a contradiction because c is irrational. Therefore, d is irrational.

To prove uniqueness we can use the fact that the addition of any two real numbers has only one result, then d is unique. Q.E.D.

$\endgroup$
  • $\begingroup$ It seem fine. But you shuold preserve the notation. And you can do your proof shorter, by assuming that x+y=z , and no mentions about integers is needed. $\endgroup$ – YTS Feb 13 '15 at 4:00
  • $\begingroup$ @YotasTrejos Now I get your idea. I am using other letters what complicate the understanding of the proof. $\endgroup$ – Beginner Feb 13 '15 at 4:28
5
$\begingroup$

You don't need to restrict the integers forming the ratio to having a greater common denominator of 1.   Just assert that they exist.


Any $z$ that is a rational number can be expressed as the ratio of two integers. (For strictness we require the denominator integer to be non-zero.)

Any $x$ is an irrational number cannot be so expressed as the ratio of two integers.

For any real numbers, $x$ and $z$, there is only a unique number $y$ where $x+y=z$.

If this $y$ were rational it could be expressed as the ratio of some two integers. For any sum $x+y$ which can also be expressed as the ratio of two integers, it would then follow that $x$ could be expressed as the sum of two integers. (By reason the product of any two integers is an integer.)

By contraposition: for any irrational $x$, and any rational $z$, the number $y$ where $x+y=z$, must be both unique and irrational.

$\endgroup$
  • $\begingroup$ Thanks for your answer! I am clear that I need to improve my writing. $\endgroup$ – Beginner Feb 13 '15 at 4:31
  • $\begingroup$ One question: At the end you wrote "by contraposition." Would we better to write by contradiction? Contraposition sounds like contrapositive, inferring “Not B implies not A” from “A implies B.” $\endgroup$ – Beginner Feb 13 '15 at 13:34
3
$\begingroup$

You have proven that the sum of a rational number and an irrational number is irrational, but you have not proven the main result.

If $x$ is irrational and $z$ is rational, what (possibly) irrational number $y$ would you wager is such that $x+y=z$? Why might this be irrational? (Hint: Use what you just proved.)

Why must it be unique?

$\endgroup$
  • 2
    $\begingroup$ Also note that you never used the fact that $\gcd(a,b)=1$, so you do not need to assume this. $\endgroup$ – user214321 Feb 13 '15 at 4:01
  • 1
    $\begingroup$ I explained what I believe you are missing. Can you tell me how you think you should modify/ expand upon your proof, given what I wrote? $\endgroup$ – user214321 Feb 13 '15 at 4:11
  • 1
    $\begingroup$ It would seem that this was covered in your discussion with Yotas Trejos. As your proof is written it only proofed a peripheral result, but when you adjust it for this context it should be fine. $\endgroup$ – user214321 Feb 13 '15 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.