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Is my proof correct?

If we define multiplication for natural numbers as

$a \times S(b) = (a \times b) + a$

$a \times 0 = 0$

And addition as

$a + 0 = a$

$a + S(b) = S(a+b)$

Where $S(n)$ is the successor function of $n$ (and assume we've already proven commutative property, etc).

Then I wish to prove that $a \times (b + c) = (a \times b) + (a \times c)$ via induction on $c$.

Base case: Let $c=0$. Then $a \times (b + 0) = a \times b$ and $(a \times b) + (a \times 0) = (a \times b) + 0 = a \times b$. Therefore $a \times (b + 0) = (a \times b) + (a \times 0)$.

Inductive Step: Suppose $a \times (b + c) = (a \times b) + (a \times c)$ for some $c$. We wish to show that $a \times (b + S(c)) = (a \times b) + (a \times S(c))$.

$a \times (b + S(c)) = a \times (S(b + c)) = a \times (b+c) + a = (a \times b) + (a \times c) + a$

and

$(a \times b) + (a \times S(c)) = (a \times b) + (a \times c) + a$

Since both sides can be shown to equal the same quantity $(a \times b) + (a \times c) + a$, we close the induction.

user525966
  • 5,957

1 Answers1

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Perfect ... though you may want to note that you need associativity of addition, since:

$a \times (b + S(c)) = a \times (S(b + c)) = (a \times (b+c)) + a = \color{red}{(}(a \times b) + (a \times c)\color{red}) + a$

while:

$(a \times b) + (a \times S(c)) = (a \times b) + \color{red}((a \times c) + a\color{red})$

Bram28
  • 103,721
  • Good catch! Additive associativity is important here. When we think of hyperoperations, I believe that the analogous "index" law for exponentiation similarly relies on associativity of multiplication, a ↑ (b + c) = (a ↑ b) * (a ↑ c). But there is no analogous law for tetration, because exponentiation is not associative. a ↑↑ (b + c) ≠ (a ↑↑ b) ↑ (a ↑↑ c) – Morgan Creighton Jul 09 '21 at 20:40