# Proving uniqueness of a mathematical object

• "Don't panic!"
In summary, the usual method for proving uniqueness of a mathematical object (for example, the identity element of a group) is to use a proof by contradiction. This involves assuming the contrary, i.e. that the object is not unique, and then showing that this leads to a contradiction. In the specific case of proving uniqueness of a mathematical object with a certain property, we assume the existence of at least one other object that also satisfies the property and then show that this other object is actually equal to the original object, hence contradicting the assumption of non-uniqueness. This shows that the original object is unique.
"Don't panic!"
As I understand it, the usual method for proving uniqueness of a mathematical object (for example the identity element of a group) is to use a proof by contradiction.
Now, for example, if we have ##a## such that ##ax=b## and we want to prove this is unique, we start by assuming the contrary, i.e. that it isn't unique. The general approach to this is to assume that there exists another element ##c## such that ##cx=b## and then show that ##a=c##. Is the reason why we only need to consider the case where one other element satisfies this property (as opposed to several other elements satisfying this same property) because if we assume that just one other element ##b## satisfies the same property as ##a##, but turns out to be equal to ##a## then we have contradicted our assumption that ##a## is not unique, in other words our assumption is false (and since this is binary logic) the contrary must be true, i.e. ##a## is unique.
Is this all there is to it? (Apologies if this is a bit convoluted)

"Don't panic!" said:
As I understand it, the usual method for proving uniqueness of a mathematical object (for example the identity element of a group) is to use a proof by contradiction.
Now, for example, if we have ##a## such that ##ax=b## and we want to prove this is unique, we start by assuming the contrary, i.e. that it isn't unique. The general approach to this is to assume that there exists another element ##c## such that ##cx=b## and then show that ##a=c##. Is the reason why we only need to consider the case where one other element satisfies this property (as opposed to several other elements satisfying this same property) because if we assume that just one other element ##b## satisfies the same property as ##a##, but turns out to be equal to ##a## then we have contradicted our assumption that ##a## is not unique, in other words our assumption is false (and since this is binary logic) the contrary must be true, i.e. ##a## is unique.
Is this all there is to it? (Apologies if this is a bit convoluted)
Yes.

No, that's not quite right (and it is somewhat convoluted). In proof of uniqueness by contradiction we don't assume that there is only one other element ## c ## that satisfies the proposition, we simply assume that there is at least one other element that is not ## a ## that satisfies the proposition (so ## c \ne a##). We then show that for any element ## c ## that satisfies the proposition, ## c = a ## is true, contradicting the assumption.

SammyS
Just to tag along with MrAnchorvy's post, when you want to prove by contradiction that x and only x satisfies some property, you assume NOT(x and only x satisfies some property) and derive a contradiction. Well, what is NOT(x and only x satisfies this property) the same as? It's the same as there is AT LEAST one other object satisfying the property.

MrAnchovy said:
No, that's not quite right (and it is somewhat convoluted). In proof of uniqueness by contradiction we don't assume that there is only one other element c c that satisfies the proposition, we simply assume that there is at least one other element that is not a a that satisfies the proposition (so c≠a c \ne a). We then show that for any element c c that satisfies the proposition, c=a c = a is true, contradicting the assumption.

So is the point that we assume that at least one other element satisfies the required property and then to carry out the proof we choose one of those elements arbitrarily and show that this other element (assumed to be not equal to ##a##) is actually equal to ##a## hence contradicting our assumption, therefore the opposite is true and the element is unique?!

pbuk
"Don't panic!" said:
So is the point that we assume that at least one other element satisfies the required property and then to carry out the proof we choose one of those elements arbitrarily and show that this other element (assumed to be not equal to ##a##) is actually equal to ##a## hence contradicting our assumption, therefore the opposite is true and the element is unique?!

Yes. It doesn't matter which one of the other objects you choose (hence the arbitrary choice), all choices lead to a contradiction.

Ok great, thanks for your help.

This doesn't need to be a proof by contradiction. You can prove it this way: let ##a## and ##b## be two objects satisfying the properties, then ##a=b##. This avoids contradiction. There is no reason to assume that ##a\neq b## in the outset.

micromass said:
This doesn't need to be a proof by contradiction. You can prove it this way: let aa and bb be two objects satisfying the properties, then a=ba=b. This avoids contradiction. There is no reason to assume that aba\neq b in the outset.

If we do this though, how can we say that this holds for any other object that satisfies the same properties as ##a## and ##b##? Is the point that we assume that we know that ##a## exists and satisfies the required properties, and then choose an arbitrary object ##b## that also satisfies these properties. If we then find that ##a=b##, then as ##b## was chosen arbitrarily, then ##a## is unique (as any other object that satisfies the required properties is equal to ##a## and so is just a different representation of the same mathematical object)?

"Don't panic!" said:
If we do this though, how can we say that this holds for any other object that satisfies the same properties as ##a## and ##b##? Is the point that we assume that we know that ##a## exists and satisfies the required properties, and then choose an arbitrary object ##b## that also satisfies these properties. If we then find that ##a=b##, then as ##b## was chosen arbitrarily, then ##a## is unique (as any other object that satisfies the required properties is equal to ##a## and so is just a different representation of the same mathematical object)?

Yes, because that is what uniqueness means: uniqueness means that if ##a## and ##b## hold the same properties, then ##a=b##. There is no inherent need for contradiction.

micromass said:
Yes, because that is what uniqueness means: uniqueness means that if aa and bb hold the same properties, then a=ba=b. There is no inherent need for contradiction.

Ah ok, I think I get it now. Just to check (apologies for the stupid question, but just want to check my understanding), is it correct then to say that it could be possible that two distinct objects satisfy the same property, but are not equal, e.g. ##a## and ##b## satisfy the same property, but ##a\neq b##, in which case the object would not be unique?

Is the point of the proof that we assume that there exists at least one other object ##b## (which may or may not be equal to ##a##, that is we don't automatically assume that ##a\neq b##) and then we proceed to show that in actual fact ##a=b##, such that ##a## and ##b## are not distinct objects, but are the same mathematical object?

Right.

Ok, thank you very much for your help!

## 1. How do you prove that a mathematical object is unique?

To prove uniqueness of a mathematical object, one must show that it is the only possible solution to a given problem or set of conditions. This can be done through logical reasoning, mathematical equations, or by providing a counterexample to any other potential solutions.

## 2. What are some common techniques used to prove uniqueness?

Some common techniques used to prove uniqueness include proof by contradiction, proof by induction, and proof by construction. These methods involve starting with a known assumption or condition and logically working through all possible solutions until the only remaining option is the unique solution.

## 3. Can a mathematical object have more than one unique solution?

No, by definition a unique solution means that there is only one possible solution that satisfies the given conditions. If there were multiple solutions, then none of them would be considered unique.

## 4. What role does mathematical rigor play in proving uniqueness?

Mathematical rigor is essential in proving uniqueness as it ensures that the solution is logically and mathematically sound. This involves using precise definitions, clear reasoning, and avoiding any assumptions or fallacies.

## 5. Are there any situations where proving uniqueness may not be necessary?

Yes, there are some cases where proving uniqueness may not be necessary. For example, if the given problem or conditions are well-defined and there is only one possible solution, then proving uniqueness may be redundant. However, in most cases, it is important to prove uniqueness to ensure the validity and reliability of the mathematical object or solution.

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