-2

This might sound dumb to a professional mathematician but why is it not useful to define log to the base 1, couldn't this help in equations just like how complex number helps?

An example: log1 3 = 3 times L.

Is there a property of the complex number that makes it natural to define it and cases like these not so much?

Bill Dubuque
  • 265,059
  • 37
  • 279
  • 908
Tangent
  • 11
  • 1
  • 4
    Because $1^x$ isn’t one to one. – Randall May 02 '22 at 17:05
  • 3
    Because of $\log_a(b)=\frac{\ln(b)}{\ln(a)}$. If $a=1$ , this leads to a division by $0$. – Peter May 02 '22 at 17:07
  • @Peter, I am not persuaded by that point. I know that (for example) in Complex Analysis, Mathematicians have conjured the idea of an *Extended Number Line* because it facilitates solving certain Math problems. This also explains why $\sqrt{-1} = i$ was *invented*. However, I suspect that the reason that logarithms base $(1)$ are not entertained is because there do not seem to be any Math problems, whose solutions would be facilitated by such conjuring. – user2661923 May 02 '22 at 17:16
  • 2
    @user2661923 Now knowing the relation to the question of defining logarithms base 1 to the question of defining division by zero... what difference do you see between this question and a question like [this one](https://math.stackexchange.com/questions/125186/why-not-to-extend-the-set-of-natural-numbers-to-make-it-closed-under-division-by)? It is not just that there are not useful math problems it solves... it is that it destroys many of the basic properties of arithmetic that we commonly rely on. – JMoravitz May 02 '22 at 17:30
  • @JMoravitz Isn't that the same point of view that Mathematicians had who initially resisted the idea of $\sqrt{-1} = i.$ Certainly, as logarithms are basically understood, yes, this inconsistency is no good. However, if Math problems do arise that benefit from creative definition of logarithm, that permits logarithms base 1, then it is feasible that such a definition could be creatively conjured that preserved consistency among the standard logarithm notions. One example of this is defining $\log x = \log |x|$, for $x < 0$. Another example of this is $\binom{n}{k} ~: n~$ is not an integer. – user2661923 May 02 '22 at 17:46
  • 1
    @user2661923 no, it is not the same. By assuming the existence of $\sqrt{-1}=i$ we do not encounter contradictions that render calculation meaningless or useless and rather encounter useful effects of its existence. Similarly, generalized binomial coefficients $\binom{n}{k}$ lead to useful effects. By assuming existence of division by zero (*and by extension, logarithms base 1*) we arrive at contradictions along the lines of $0=1$, effectively showing that there is only a single number in our entire number system if we allow that, which makes arithmetic involving it useless. – JMoravitz May 02 '22 at 17:50
  • @user2661923 **"Isn't that the same point of view that Mathematicians had who initially resisted the idea of $\sqrt{-1}=i.$"** No, it is not. Mathematicians opposed the use of complex numbers on philosophical grounds, not mathematical grounds. There is no contradiction that emerges from guaranteeing the existence of some complex number $x$ satisfying $x^2=-1.$ Mathematicians just did not like the idea that numbers should be two-dimensional. It makes sense, since the idea of a vector space did not yet exist during the time. – Angel May 02 '22 at 19:30
  • 1
    @user2661923 There is no way of finding a number system where division by $0$ is possible. This is not even a matter of utility. There is a theorem in abstract algebra that prevents multiplication by $0$ from being cancellable. The theorem is as follows: suppose $(M,+)$ and $(M,\cdot)$ are magmas, and suppose $0+x=x+0=x$ and $x\cdot(y+z)=x\cdot{y}+x\cdot{z}$ and $(y+z)\cdot{x}=y\cdot{x}+z\cdot{x}.$ Then if, for some given $a,$ if $0\cdot{a}$ is $+$-cancellable, then $0$ is not $\cdot$-cancellable. – Angel May 02 '22 at 19:34
  • 1
    Does this answer your question? [Why don't we define "imaginary" numbers for every "impossibility"?](https://math.stackexchange.com/questions/259584/why-dont-we-define-imaginary-numbers-for-every-impossibility) – Hans Lundmark May 02 '22 at 20:52

1 Answers1

3

To understand the answer to your question, I think it is important to begin by first understanding what is a logarithm, in general. What is a logarithm? Well, a logarithm is a function satisfying a certain important property. There are many ways to define a logarithm, but the modern way to define logarithms that takes into account the history of them is by looking at the equation $f(xy)=f(x)+f(y).$ This equation just encodes the fact that logarithms convert multiplication into addition. That is the reason they were developed, historically. Logarithms are the continuous functions satisfying this property. The logarithm with the property that $f(b)=1$ is called the logarithm with base $b.$

So what gives? Well, it can be proven that every logarithm has the property that $f(1)=0.$ You can conclude this from just using the equation $f(xy)=f(x)+f(y).$ So there is no logarithm satisfying $f(1)=1,$ and as such, there is no logarithm with base $1.$ Any function satisfying $f(1)=1$ cannot satisfy the equation $f(xy)=f(x)+f(y).$ So this is not a problem you can solve by just introducing a new number system.

Angel
  • 3,606
  • 11
  • 13