As stated in the title, I'd like to find out is whether or not it is always mathematically sound to prove the uniqueness of something before proving the existence of said something.

I am still relatively new to proofs and, in this light, I was intrigued by an observation that some authors choose to prove uniqueness before existence. I understand that some do this because, allegedly, "it is easier to prove the uniqueness part first". And in an exam setting, it's better to tackle the easier parts of a task.

However, as I see it, if done this way, aren't we essentially proving a proposition about something that, at that point, we have not yet guaranteed exists? (Thus, might not even exist at all). I understand that in most cases the existence is proved immediately after the proof of uniqueness but, generally, doesn't uniqueness depend on existence?

My example will be derived from page $40$ of The Real Analysis Lifesaver: All the Tools You Need to Understand Proofs by Raffi Grinberg:

Theorem 5.8. (Existence of Roots in $\mathbb{R}$)

Every positive real number has a unique positive $n$th root, for any $n \in \mathbb{N}$.

In symbols:

$∀x ∈ R$ with $x > 0, ∀n ∈ N, ∃y ∈ R$ unique, such that $y > 0$ and $y^n = x.$

Note that even-numbered roots $(\sqrt{x},\sqrt[4]x,\sqrt[6]x, etc.)$ signify two numbers in $\mathbb{R}$, namely, $+y$ and $−y$. The theorem asserts that there is one and only one positive real root.


The uniqueness of $y$ is the easiest part of the proof, so we’ll start there. For any two different positive real numbers, the fact that they are different means one must be greater than the other. If there were two positive real roots $y_1$ and $y_2$ such that $y_1^n = x$ and $y_2^n = x$ we would have $0 < y_1 < y_2$. But then $0 < y_1^n < y_n^2$ , meaning $0 < x < x$, which is impossible. Thus only one positive real root can exist.

To prove that $\sqrt[n]x$ exists in $\mathbb{R}$, let’s first figure out our game plan and then write it up formally $$...$$

While the proof is totally clear, I'd really like feedback on this format. As an aspiring mathematician, is this a format I can adapt and use in my proofs or perhaps it is not good practice in general? What if, say, I have a conjecture that there exist unique integers possessing some property and this conjecture is something new (i.e. hasn't been proven yet). I try proving they exist but run out of luck. Would it be worth the effort of trying to prove the uniqueness of such integers not knowing they exist?

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    $\begingroup$ Yeah, that's perfectly fine. It doesn't matter: you're proving two things: 1) There is at least one of these; and 2) there aren't two different ones. There's no problem doing them whatever way round you like. There are a fair few results of the form "there are at most $n$ things with property X", and this is just the special case where $n = 1$. $\endgroup$ – user3482749 Jan 4 at 17:57
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    $\begingroup$ Sure. Naturally, if you show uniqueness first what you have actually shown at that point is that there is at most one object with the desired property. $\endgroup$ – Andrés E. Caicedo Jan 4 at 17:57
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    $\begingroup$ Even in the case where authors first prove uniqueness, they usually mean, under the assumption that there exists such an object, it is unique. $\endgroup$ – Dietrich Burde Jan 4 at 18:01
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    $\begingroup$ I prefer existence before uniqueness, but that is just a preference and not a strong recommendation. $\endgroup$ – David G. Stork Jan 4 at 18:02
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    $\begingroup$ An explanation: I prefer existence before uniqueness because in general existence is simpler. Imagine the awkward case where you prove uniqueness first, then find you cannot prove existence! THAT is weird. $\endgroup$ – David G. Stork Jan 4 at 18:16

I can see the unease: if what you're proving unique doesn't exist, might that invalidate whatever manipulations you've done with it?

This is similar to the situation with proof by contradiction, where you don't just reason about something that might not be true: you reason about it expecting it to be untrue, but making all the logical steps anyway.

In either case, you end up showing "If $A$ then $B$"—whether it's "If $x$ exists then it's unique" or "If $x$ exists then it has contradictory properties". In one case we go on to apply $B$ to $x$ once we know it exists, and in the other we use the impossibility of $B$ to disprove the existence of $x$.

A separate thought: proofs are presented as a sequence of steps and inferences because as humans, we process written information sequentially. But what's behind a mathematical theorem is more like a network of theorems and logical steps, some dependent on others, which together imply the theorem. We can't take them all in simultaneously, so we arrange them in a suitable order to let us see the logic most clearly. It's a matter of convenience.

If $X$ and $Y$ are both true and together imply $Z$, it doesn't really matter which order the various parts of that are proved in—all that matters from a validity point of view is that they're all proved and brought together.

  • $\begingroup$ Thanks! Your first paragraph alone asks my question even better than I could in my long post! $\endgroup$ – E.Nole Jan 7 at 2:56

This is a perfectly reasonable argument. The author even tells you he's doing it this way. He then uses the argument to allow the reader to become comfortable with the assertion and the notation.

There are other reasons for starting with uniqueness. Often if you know there's just one of something then it might be easier to search for it.

When faced with a problem with an easy part and a hard part the argument for starting with the easy part is that it's a good warmup. That's what I usually do, and what I like to see in things I read.

The argument for starting with the hard part is that if you fail there you haven't wasted time on the easy part. For example, proving that there's at most one positive rational number whose square is $2$ is not going to tell you very much about the rational square root of $2$ that does not exist.

Starting with the hard part, the bottleneck, is often the preferred strategy when writing software.

PS The author is wrong to say that when $x >0$, $\sqrt{x}$ signifies two two numbers. $4$ does have two real square roots, but the mathematical convention is that only $2 = \sqrt{4}$.

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    $\begingroup$ Nice typo near the end: "two two" (Shame to fix it really!) $\endgroup$ – timtfj Jan 4 at 19:15
  • $\begingroup$ @timtfj I will leave it thanks. $\endgroup$ – Ethan Bolker Jan 4 at 19:33

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