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It seems that there is a marked discrepancy between how one might explain a concept in mathematics, and how one might prove it. Obviously, there are some exceptions, and some particularly elegant proofs merely formalise what we intuitively know. However, more often than not, formal rigour seems to be at the cost of understanding something in a more 'human' way.

To illustrate, here is an elementary example:

Prove $\log_c c^x=x$

Here, my attempt at a formal proof would be:

\begin{align} \log_c c^x &= x\log_c c \text{ (Using the power law)} \\ \log_c c&=1 \text{ as $c^1=c$ by definition} \\ \implies x\log_cc&=x(1)=x \end{align}

I wonder if this proof might be seen as verbose, or overly formal. Would the following also be accepted as a proof?

\begin{align} \log_c c^x &= x\log_c c=x \end{align}

On the other hand, perhaps a mathematician might say that neither of these proofs are formal enough, and that to properly prove $\log_cc^x=x$, one must lay out all of the axioms/definitions that they are using, and build the proof from the ground up.

And finally, here is the way I intuitively understand $\log_c c^x$:

$\log_ca$ means 'to what power must $c$ be raised in order to get $a$?'. Therefore, in the case of $\log_cc^x$, the question becomes 'what power must I raise $c$ to in order to get $c^x$?'. Therefore, the question is easy to answer. It's like writing $c^?=c^x$, where the answer to the question is very obviously $x$.

Unfortunately, I see no way of turning this explanation into a rigorous proof. Perhaps it is already, if you modify it slightly, a rigorous proof.

Joe
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    Unfortunately, your example is not the good one... $\log_c c^x=x$ by definition of $\log$. – Mauro ALLEGRANZA Jun 25 '20 at 12:15
  • I would have said that $\log_c c^x=x$ was a definition, not a theorem. whichever one you have defined previously, $\log_c z$ or $c^z$, the other is defined as the inverse function. – lulu Jun 25 '20 at 12:15
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    IMO, "rigorous" does not mean "avoid human language". It means avoid misunderstanding, avoid unnoticed assumptions, etc. – Mauro ALLEGRANZA Jun 25 '20 at 12:17
  • It can't get anymore formal than: $c^?$. Jokes aside, I would consider it formal, just replace 'means "to what..."' with: $\log_b c=y \iff c=b^y$ – Bertrand Wittgenstein's Ghost Jun 25 '20 at 12:30
  • $\log_c c^x=y \iff c^y=c^x$ Therefore, y=x=$\log_c c^x$ – Bertrand Wittgenstein's Ghost Jun 25 '20 at 12:31
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    Besides the mathematical error in your proof, as explained in other comments, your perception of formality seems somewhat limited to me. Human language expressions can be just as formal as symbolic expressions. I'm sure you'll agree that there is utterly no difference between "this $\implies$ that" and "this implies that" and "this is sufficient for that" and "that is necessary for this". And, of course, it goes on much further and deeper into other expressions of logic. This is a matter of training and experience more than anything else. – Lee Mosher Jun 25 '20 at 12:48
  • Related: https://math.stackexchange.com/questions/3084938/when-is-something-considered-to-be-a-mathematical-proof/3084949#3084949 – Ethan Bolker Jun 25 '20 at 15:25

1 Answers1

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Here is a way to turn your last intuitive argument into a formal proof ^_^

1 The definition of $\log$

The definition of $\log_ab$ is that it is the number $x$ which you raise $a$ to get $b$.

$$\log_a b = x \quad\equiv\quad a^x = b $$

Taking $x$ to be $\log_a b$ makes the right side true and the left side becomes: $$a^{\log_a b} = b$$

That is, raising $a$ by $\log_a b$ gives $b$ ---which is the desired interface for $\log$.

2 The proof

It is obviously clear that to get $c^x$ from $c$, you just raise $c$ by $x$.That is:

\begin{align} & \log_c (c^x) = x \\ \equiv\quad & c ^ x = c^x & \{\text{ Definition of $\log$ }\} \\ \equiv\quad & \mathsf{true} & \{\text{ reflexivity of equality }\} \end{align}

2 Discovering a Solution

It is not so obviously clear how to get $b^c$ from $a$. Rather than guess and check, let us calculate! Let $x$ be the unknown solution which we are look for.

\begin{align} & \log_a (b^c) = x \\ \equiv\quad & a^x = b^c & \{\text{ Definition of $\log$ }\} \\ \equiv\quad & a^x = (a^{\log_a b})^c & \{\text{ To get to $b$ from $a$, use $\log_ab$ }\} \\ \equiv\quad & a^x = a^{((\log_a b) \cdot c)} & \{\text{ Exponents }\} \\ \equiv\quad & x = ((\log_a b) \cdot c) & \{\text{ Exponents }\} \\ \equiv\quad & x = c \cdot \log_a b & \{\text{Arithmetic }\} \end{align}

Hence, we have calculated the power rule for logarithms.

There were some jumps in the above calculation, and so it could be massively improved, but it's in the right spirit of calculating from definitions.

Musa Al-hassy
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    I just want to say that $\log_a(b)=x\iff b=a^x$ is ONE definition of $\log$, not THE definition. You could just as well define it by $\log_a(a^x)=x$. If you wanted, you could define $\log$ by its Taylor series. There's this idea in modern mathematics that formal rigorous proof is good, but hundreds of years ago people would publish their discoveries in the form of challenges or riddles. That system was probably worse, but the point is it's all subjective. – A. Kriegman Jun 25 '20 at 15:31
  • Yours is not a formal proof at all!! It fails for $c = -2$ and $x=4$. Please, don't claim to have a formal proof if you don't. – user21820 Jun 25 '20 at 15:34
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    I'm pretty sure it is still a correct proof even when $c$ is negative, but either way there is an implicit assumption that $c$ is positive because you need to bring in complex numbers to have a logarithm with a negative base. – A. Kriegman Jun 25 '20 at 15:47