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Prove that the following operator is associative for $b\in \Bbb N$

$$x||y = x\cdot b^{1+\lfloor\log_{b}{y}\rfloor}+y$$

One thing that you can notice is that it is the concatenation operator. However, you are not allowed to use this fact. In first order logic, we refuse to attach any meaning to any objects and try to prove things starting with axioms.

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    "One thing that you can notice is that it is the concatenation operator." Since this looks like it's likely to be a homework problem that you've copied and pasted, first things first: Do you understand why this is the case? – 2017-02-07
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    @DustanLevenstein This is not a homework question. I watched a youtube video which claimed concatenation to be unmathematical. So I came up with a formula to represent it. Now I am curious if all the properties of concatenation can be rigorously proved. – 2017-02-07
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    In that case, +1. I don't think you're likely to receive an answer though. – 2017-02-07
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    I'm not sure if the operator is exactly how you intend. Pick b = 10. Then $1 || 1= 1 * 10^0 + 1 = 2$. And $10 || 10 = 10*10 + 10 = 110$. So this operator isn't really concatenation – 2017-02-13
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    Perhaps you wanted $$x||y = x\cdot b^{\lfloor\log_{b}{y}\rfloor + 1}+y$$?? – 2017-02-13
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    @BadamBaplan - That's right. The OP missed out on a '$+1$'. https://en.wikipedia.org/wiki/Concatenation_(mathematics)#Calculation – 2017-02-13
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    The idea that "concatenation is unmathematical" is so absurd it makes me dizzy. You should watch better YouTube videos... – 2017-02-13
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    @MarianoSuárez-Álvarez Have you seen https://www.youtube.com/watch?v=Noo4lN-vSvw ? – 2017-02-14

1 Answers 1

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I will be working off the formula $$(x || y) = x * b^{\lfloor\log_b(y)\rfloor + 1} + y$$ You can see that this should be the case because with $b = 10$, $y \in [10,99]$ should multiply $x$ by $100$.

In the following, I will assume $b = 10$ and write $\log(x)$ for $\log_{10}(x)$. (The proof generalizes immediately to any $b$). The proof does not require case analysis, only applying a couple elementary properties of the floor function.

The more interesting point is perhaps about whether concatenation is 'unmathematical.' I'm not sure what you mean by 'unmathematical'. In general we want to be able to define concatenation of words for arbitrary symbolic systems (alphabets). I suppose you (or this youtube poster) are taking issue with the dependency on indexing (and knowing the length of the second argument) in the usual definition of concatenation.

We have by definition of the operator $||$

$\big(x || y \big) = 10x * 10^{\lfloor\log(y)\rfloor} + y$

$\big(y || z\big) = 10y * 10^{\lfloor\log(z)\rfloor} + z$

So $\big(x || y\big) || z =$ $$10*\big(x || y\big)*10^{\lfloor\log(z)\rfloor} + z =$$ $$(100x * 10^{\lfloor\log(y)\rfloor} + 10y)*10^{\lfloor\log(z)\rfloor} + z = $$

$$100x*10^{\lfloor\log(y)\rfloor + \lfloor\log(z)\rfloor} + 10y*10^{\lfloor\log(z)\rfloor} + z$$

Meanwhile $x||\big(y||z\big) = $ $$x || \big(10y * 10^{\lfloor\log(z)\rfloor} + z\big) =$$ $$10x * 10^{\lfloor \log (10y * 10^{\lfloor\log(z)\rfloor}) \rfloor} + 10y * 10^{\lfloor\log(z)\rfloor} + z = $$ $$10x * 10^{\lfloor 1 + \log(y) + \lfloor\log(z)\rfloor \rfloor} + 10y * 10^{\lfloor\log(z)\rfloor} + z$$

Consider that $\lfloor 1 + a + \lfloor b \rfloor \rfloor = 1 + \lfloor b \rfloor + \lfloor a \rfloor$, since $1 + \lfloor b \rfloor$ is an integer. Applying this to the last line in the expansion of $x||\big(y||z\big)$ shows $x||\big(y||z\big) = $ $$10x * 10^{1 + \lfloor \log(y) \rfloor + \lfloor\log(z)\rfloor} + 10y * 10^{\lfloor\log(z)\rfloor} + z = $$ $$100x * 10^{\lfloor \log(y) \rfloor + \lfloor\log(z)\rfloor} + 10y * 10^{\lfloor\log(z)\rfloor} + z$$.

This shows that $x||\big(y||z\big) = \big(x||y\big)||z$ as desired, and we have proved that your formal "concatenation" operator satisfies associativity!

Now, have we actually mathematically captured concatenation?

The only problem to get around is zeros. $x || 0$ is not well-defined for us, and $0 || y$ returns $y$, so it isn't really concatenation in a string sense. Moreover if a number contains $0$ in it's middle, then treating it as a word we see that is the concatenation of $x||y$ where $y$ has a leading zero, and the formula breaks. The problems are that our formula deals with numbers, but concatenation deals with strings. In the former, leading zeros don't matter, but in the latter, they do.

This isn't ultimately a problem though. We can use this formula to define concatenation for any finite symbolic system.

Let $||$ be defined as above on any base $b$ number system with the explicit definition $x || 0 = x$ for all $x$. Given an alphabet $A$ consisting of $n$ characters, we define concatenation on two words $u,v$ of $A$ as follows: $$u || v = \phi^{-1}\big(\phi(u) || \phi(v)\big)$$ where $\phi$ is any bijection $\phi: A \rightarrow \{1,\ldots,n\}$ extended to act on words of $A$ elementwise, with its image on the empty word explicitly defined to be $0$. Note that $\phi$ is then a bijection between words of $A$ and numbers base $n+1$ which either are zero or have no zeros as digits.

So the operator of concatenation on words of any alphabet can be defined quite rigorously without making reference to the lengths or elements of the words. We only require for this that the alphabet be finite, because our formula for concatenation in a base $b$ number system breaks down for $b$ infinite.