5
$\begingroup$

One day, I noticed that

$2+2 = 2*2 = 4$.

Later, I learned that

$2+2 = 2*2 = 2^2 = 4$

Multiplication is an abstraction of a lot of additions , exponential is an abstraction of a lot of multiplications... I'm sure there is always an abstraction of the previous operation.

But my questions is : if I use any abstraction with the numbers $2$ and $2$ does it always result $4$ ?

$\endgroup$
5
  • 7
    $\begingroup$ $2/2=1$, case closed. $\endgroup$
    – Raskolnikov
    Commented Sep 4, 2013 at 13:10
  • 4
    $\begingroup$ Look at en.wikipedia.org/wiki/Hyperoperation and at en.wikipedia.org/wiki/Ackermann_function. $\endgroup$
    – Soham Chowdhury
    Commented Sep 4, 2013 at 13:11
  • 1
    $\begingroup$ @Raskolnikov $3/3 = 1$ and $3+3 ≠ 3^3$ I don't understand what you mean. $\endgroup$
    – Pyrofoux
    Commented Sep 4, 2013 at 13:13
  • 4
    $\begingroup$ I don't understand the downvotes. The question is asking whether $2$ tetrated to $2$ or $2$ pentated to $2$ and so forth always results in $4$ (and also a proof it seems). Seems like a valid question to me. $\endgroup$
    – Alraxite
    Commented Sep 4, 2013 at 13:21
  • $\begingroup$ See Corollary 2(iii) on p. 97 of: John Doner and Alfred Tarski, An extended arithmetic of ordinal numbers, Fundamenta Mathematica 65 (1969), 95-127. $\endgroup$
    – Dave L. Renfro
    Commented Sep 4, 2013 at 16:46

1 Answer 1

Reset to default
10
$\begingroup$

I think the "deep" reason for this is that these are all binary operations.

Given a binary operation $\ast$ on integers at least $2$, define $\ast'$ by $$m\ast' n = \overbrace{m\ast m\ast \cdots \ast m}^{n\text{ times}}.$$ Always associate to the right (for concreteness), so that $a\ast b\ast c = a\ast(b\ast c)$ and so on.

Now starting with any binary operation $\ast$, define $\ast_1=\ast$ and $\ast_{n+1} = \ast_n'$. Then for all $n$ we clearly have $2\ast_n 2 = 2\ast_{n-1} 2 = \cdots = 2\ast 2$.

The sequence you consider is given by taking $\ast$ to be $+$.

$\endgroup$
6
  • $\begingroup$ This is what we always do in a vector space. $\endgroup$
    – Mikasa
    Commented Sep 4, 2013 at 13:31
  • 7
    $\begingroup$ Exercise: Find $\ast$ such that $\ast'$ is $+$. $\endgroup$
    – Sean Eberhard
    Commented Sep 4, 2013 at 13:37
  • $\begingroup$ @SeanEberhard I think the operation $*$ such that $*'$ is $+$, is incrementation and needs just one number. :) $\endgroup$
    – Pyrofoux
    Commented Sep 4, 2013 at 14:17
  • $\begingroup$ @SeanEberhard: Well, as $a\ast'1=a$, that makes the exercise tricky... $\endgroup$
    – George Lowther
    Commented Sep 4, 2013 at 15:29
  • $\begingroup$ @GeorgeLowther I only defined $\ast'$ on integers at least $2$. $\endgroup$
    – Sean Eberhard
    Commented Sep 4, 2013 at 15:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .