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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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