Is there such a function $f(a,b)=k$ where each value of $k$ appears only once for all integer values of $a$ and $b$?

Question

Suppose we have $f(a,b)=k$ such that $k$ is an integer when $a$ and $b$ are integers. Is there such a function where each value of $k$ appears only once for all integer values of $a$ and $b$?

e.g the function $f(a,b)=4ab+3a+2b$ satisfies the qualities for a and b between $0$ and $2$ inclusive.

I can find functions for small ranges of $a$ and $b$, but I have yet to find one that satisfies the property for all integers. Any thoughts?

@B.Goddard I didn't even think about exponential equations, I'm a fantastic moron — , Mar 12 '18 at 21:01
@B.Goddard: that doesn't work for all integers $a,b$, because $2^a3^b$ isn't an integer for negative $a,b$. — symplectomorphic, Mar 12 '18 at 21:07
@symplectomorphic It's not clear to me that the OP wanted an onto function. If so, we also have to worry about all the other primes. — B. Goddard, Mar 12 '18 at 21:09
It's not about being onto: your function isn't a map $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$. The OP didn't say positive integers. — symplectomorphic, Mar 12 '18 at 21:09
@mcchucklezz But "only once" is not the same as "exactly once." So I took it to mean "Once or nonce." — B. Goddard, Mar 12 '18 at 21:23
@B.Goddard: you are straight up misreading. The issue is not surjectivity; it's that your function doesn't have the right domain and co-domain, regardless of the image. — symplectomorphic, Mar 12 '18 at 21:25

symplectomorphic · Answer 1 · 2018-03-12T21:33:30.870

I can't quite tell whether you're asking for an injection or a bijection $f:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$. (An injection guarantees that each $k$ is output only once, if it is output at all; but it may not be output unless we in fact have a bijection.)

There are infinitely many such functions, but here's a fun and rather natural injection. It has a "computational" flavor: unlike mysterious rabbit-out-of-the-hat Cantor factorizations, it is the sort of thing a programmer or cryptographer might think of.

To get an injection, we need to be able to recover the inputs based only on the output. So why not directly encode the inputs into the output? Given integers $a$ and $b$, take their binary numerals; interpret them as ternary numerals instead; then concatenate them with an intervening digit of $\mathtt{2}$. Then let $f(a,b)$ be the integer corresponding to that concatenated numeral. What about negatives? If $a$ is negative, tack on a $\mathtt{2}$ at the left of its binary numeral before concatenating; if $b$ is negative, tack on a $\mathtt{2}$ to the right of its binary numeral before concatenating. This way the ternary numeral of the output will always have at most three $\mathtt{2}$'s: one in the middle to separate the binary numerals of the inputs, and possibly two more at the edges, depending on whether the inputs are negative.

For example, $f(8,11)=6754$ because $8$ is $\mathtt{1000}$ in binary, and $11$ is $\mathtt{1011}$ in binary, so their concatenation according to the above scheme is $\mathtt{100021011}$, which represents the integer $6754$ in ternary. Similarly, $f(-1,3)=211$ because $1$ is $\mathtt{1}$ in binary, and we append a $\mathtt{2}$ at the front to get $\mathtt{21}$ before concatenating; $3$ is $\mathtt{11}$ in binary; so their concatenation is $\mathtt{21211}$, which represents the integer $211$.

Cantor's not really a rabbit out of a hat: it's just walking through the plane in diagonals. — Patrick Stevens, Mar 14 '18 at 07:39
Yes, of course, but it's the idea of walking through in diagonals that's a non-rabbit out of a non-hat, not the bald assertion of the final formula. :) In any case, I told a white lie to emphasize there are very different ways to think about the problem. — symplectomorphic, Mar 14 '18 at 16:04

score 2 · Accepted Answer · answered Mar 12 '18 at 21:09

2

There's a standard bijection, the Cantor pairing function, which takes $(a, b) \mapsto \frac{1}{2} (a+b)(a+b+1) + b$. This is a bijection $\mathbb{N} \times \mathbb{N} \to \mathbb{N}$.

To create a bijection $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$, just take your favourite bijection $\mathbb{Z} \to \mathbb{N}$ (e.g. if $n$ is nonpositive, send it to $-2n$, otherwise to $2n-1$), and compose with that.

answered Mar 12 '18 at 21:09

Patrick Stevens

36,135

I'll have to do some reading on these before I accept the answer, but what I do understand makes sense in my case. Thanks! – Mar 12 '18 at 21:13

Is there such a function $f(a,b)=k$ where each value of $k$ appears only once for all integer values of $a$ and $b$?

2 Answers2