# Proving uniqueness of a mathematical object

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1. Jul 2, 2015

### "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)

2. Jul 2, 2015

Yes.

3. Jul 2, 2015

### MrAnchovy

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.

4. Jul 3, 2015

### JonnyG

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.

5. Jul 3, 2015

### "Don't panic!"

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?!

6. Jul 3, 2015

### JonnyG

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

7. Jul 3, 2015

### "Don't panic!"

Ok great, thanks for your help.

8. Jul 3, 2015

### micromass

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.

9. Jul 3, 2015

### "Don't panic!"

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)?

10. Jul 3, 2015

### micromass

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.

11. Jul 3, 2015

### "Don't panic!"

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?

12. Jul 3, 2015

### micromass

Right.

13. Jul 3, 2015

### "Don't panic!"

Ok, thank you very much for your help!