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When proving that there is a unique mathematical object that satisfies a particular condition, e.g., the inverse of an element of a group, is the intuition behind it the following?

You assume that the solution set (i.e., the set of objects that satisfy this condition) contains an arbitrary number of elements and then you choose two elements arbitrarily from this set. Then, if after a set of logical steps you are able to show that these two elements are actually equal, then (by the transitive property of equality) all elements of the solution set are equal to one another and hence there is actually only one unique solution.

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    $\begingroup$ I see a question mark but no question. $\endgroup$ – Matt Samuel Feb 21 '15 at 18:22
  • $\begingroup$ Yes, $\ |S|\ \le 1\iff \forall x,y\in S\!:\ x = y\ \ $ $\endgroup$ – Bill Dubuque Feb 21 '15 at 18:24
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    $\begingroup$ You have a low answer-acceptance rate. Please read about accepting answers here and here. $\endgroup$ – Git Gud Feb 21 '15 at 18:28
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    $\begingroup$ Seems to be essentially the same as your question 5 months ago, and your first question. $\endgroup$ – Bill Dubuque Feb 21 '15 at 18:29
  • $\begingroup$ Sorry, I know I've asked it before, but I felt I never got full clarification from the early post. $\endgroup$ – Will Feb 21 '15 at 18:55
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Yes, you have correctly outlined the general process to show uniqueness.

Just be careful that you must also show existence. Sometimes the existence of a thing is clear or easy to construct; in your example of inverses of group elements, one of the properties of a group is that every element has an inverse, so you're covered there.

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