I have to prove this proposition: Given $m,n \in\mathbb Z$, there exists one and only one $x \in\mathbb Z$ such that $m + x = n$. So, just to be sure: I am given an equation and asked to first prove the existence of solution x and then its uniqueness, right?

From what I understand, "uniqueness" simply means that if two solutions fit in the equation, they must be equal. I can only use the axioms for addition and multiplication and derived propositions.

Proof of existence:

\begin{align*} m + x &= n\\ (-m) + (m + x) &= (-m) + n\\ (-m + m) + x &= (-m) + n\\ 0 + x &= (-m) + n\\ x &= (-m) + n\\ \end{align*}

Hence, the solution x exists.

The second part seems too simple. Assuming that x can take two values, x1 and x2... Proof of uniqueness: \begin{align*} m + x1 &= n\\ m + x2 &= n\\ m + x1 &= m + x2\\ (-m) + (m + x1) &= (-m) + (m + x2)\\ (-m + m) + x1 &= (-m + m) + x2\\ 0 + x1 &= 0 + x2\\ x1 &= x2 \end{align*}

Hence, the solution x is unique. What do you think? Thank you!


Suppose $m+x_1=n$ and $m+x_2=n$. Then $m+x_1=m+x_2$ and so $x_1=x_2$ by left-cancellation. I think you did a great deal more work than you really needed to here, but it never hurts to try to be thorough.

| cite | improve this answer | |
  • $\begingroup$ I should note that this, of course, does not address the existence portion of your question; for that, you need to make use of whatever axioms are at your disposal (just as I used left-cancellation because I assumed you could use it). $\endgroup$ – Daniel W. Farlow Feb 7 '15 at 2:54
  • $\begingroup$ But this is OP's proof, just replacing adding $-m$ to both sides and using associativity and $m+(-m)=0$ with the words 'by left-cancellation'. This is not something you should omit given that OP said "I can only use the axioms for addition and multiplication and derived propositions". $\endgroup$ – user26486 Feb 8 '15 at 3:11
  • $\begingroup$ @user314 This is far more succinct. I mean the question really is meaningless without knowing what axioms or results you're allowed to use. $\endgroup$ – Daniel W. Farlow Feb 8 '15 at 3:13

In your proof of existence you basically show that the equation $$m+x=n$$ is equivalent to the equation $$x=(-m)+n$$ But it is clear that this last equation has one unique solution. Hence you have actually already proved the uniqueness.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your input. I am new to proofs, so why would x = (-m) + n also show uniqueness? Because a binary operation can only generate one element? $\endgroup$ – Johnathan Feb 8 '15 at 21:49
  • $\begingroup$ You have on the left hand side your unknown variable, and on the right hand side a well defined number. Such an equation has always a unique solution, which is given by the right hand side. $\endgroup$ – Gilles Bonnet Feb 9 '15 at 11:10
  • $\begingroup$ And yes, the binary operation $(a,b)\mapsto a+b$ goes from $\mathbb{Z}\times\mathbb{Z}$ to $\mathbb{Z}$. So you can say that it generates one element. $\endgroup$ – Gilles Bonnet Feb 9 '15 at 11:12

You did a great job with your proof! Why?

Your strategy is in the page 58 of the edition 2014 of the book "A Transition to Advanced Mathematics", written by Smith, Eggen and Andre.

"PROOF OF (∃!x)P(x): (i) Prove that (∃x) P(x) is true. Use any method. (ii) Prove that (∀y)(∀z)[P(y) ∧ P(z) ⇒ y = z]. Assume that y and z are objects in the universe such that P(y) and P(z) are true. From (i) and (ii) conclude that (∃!x)P(x) is true."

My proof would be:

Assume m, n, x, y, and z are integers; and m + x = n.

There exists a solution because:

-m + m + x = -m + n (-m is the inverse of m under addition.)

x = -m + n (-m + m = 0 because 0 is the identity element under addition.)

x = n - m (commutative property).

As a final step we will be prove the uniqueness of the solution.

Assume we have two differents solutions y and z.

y = n - m, and z = n - m


y = z.

For that reason, the solution is unique. Q.E.D.

| cite | improve this answer | |
  • $\begingroup$ @Johnathan Please give me your opinion about my new answer to your question "Prove the uniqueness of subtraction." I am learning to write answers. $\endgroup$ – Beginner Feb 9 '15 at 5:21

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.