[Math] Ackermann function proof by Induction

ackermann-functiondiscrete mathematicsinductionrecursion

I'm currently studying discrete mathematics and i've been given an assignment to prove the following: $A(1, n) = n +2$ for all $ n \geq 0$ with induction. But i am somewhat unsure if i've done it correctly.

Ackermanns function is defined as:

$A(0, n) = n + 1, n \geq 0$
$A(m, 0) = A(m – 1,1) ,m > 0$
$A(m, n) = A(m – 1, A(m, n – 1)), m, n > 0 $

Here's what i've done so far:

Base Case, $n = 0$: $A(1,0) = A(1-1, 1) = A(0,1) = 1+1 = 2$

Induction Hypothesis, $n = k$: Assume true for $A(1, k) = k + 2$, when $k \geq 0$

Prove: $n = k + 1$:

I followed the third function because if we have $k + 1$ then $k > 0$

(1): $A(1, k + 1) = A[1-1, A(1, k+1-1)]$

Question: Am i allowed to set $k = 0$ when proving n = k + 1?

If $k = 0$

Calculate the inner function of (1): $A(1, k+1-1) = A(1,0)$

(2): $A(1, 0) = A(1-1,1)$

Calculate function from (2): $A(1-1,1) = A(0, 1)$

(3): $A(0, 1) = 1 + 1 = 2$

Replace the inner function of (1) with the result from (3)

(1): $A(1, 0 + 1) = A(0, 2) = 2 + 1 = 3$

Best Answer

The induction is on $n$.

Your first step is correct; for $n=0$ we have that:

$A(1,0)=A(0,1)=1+1=2$,

that is $n+2$ for $n=0$.

For the induction step, we have to assume that the property holds for $k$ and prove it for $k+1$.

Thus, we assume the induction hypotheses: $A(1,k)=k+2$, and we have to compute:

$A(1,k+1)=A(0,k+2)=k+3$,

which is $n+2$ for $n=k+1$.

Related Solutions

Discrete Mathematics – Concept of Mathematical Induction

What is induction?

At its core, induction is this statement:

Take a set $S$. If $1 \in S$, and $k \in S$ implies that $k + 1 \in S$, then all natural numbers are in $S$.

Intuitively, this is pretty obvious. Assume both hypotheses: that $1 \in S$ and $k \in S \implies k + 1 \in S$.

Is $1 \in S$? Yes, by assumption.
Is $2 \in S$? Yes, because $1 \in S$, and $1 \in S$ implies $1 + 1 \in S$.
Is $3 \in S$? Yes, because we just showed $2 \in S$, and $2 \in S$ implies $2 + 1 \in S$.
$\ldots$

That clearly wasn't a formal proof that induction is valid, but it's a good place to start your intuition. No matter what natural number we start at, we can "step back" by one until we hit $1$, which we know is true.

Wait, how'd sets get in there?

Normally, people don't talk about sets when it comes to induction. Usually, it's about propositions. But the two are equivalent, consider the set $S = \{ n \mid P(n) \}$ or the proposition $P(n) = n \in S$. If you're working with propositions, induction is stated as:

Take a proposition $P$. If $P(1)$ is true, and $P(k)$ implies that $P(k+1)$, then $P(n)$ is true for all $n \in \mathbb{N}$.

How do we do inductive proofs?

Like any conditional statement, all you need to do is prove the two hypotheses:

$1 \in S$ (the base case)
If $k \in S$, then $k + 1 \in S$ (the recursive case)

or for propositions,

$P(1)$ (the base case)
If $P(k)$, then $P(k + 1)$ (the recursive case)

The confusing part

The recursive step sometimes confuses people. "Aren't we assuming what we want to prove?" Not quite. Here, a precise understanding of conditional statements is required. If I say, "If today is Friday, tomorrow is Saturday", I am not saying that today is Friday. I am only saying that, if we can show today is Friday, then we know tomorrow is Saturday.

It is the same with the recursive step. We are proving: "if $P$ is true for $k$, then it is also true for $k + 1$". We have not said anything about whether or not $P(k)$ is true!

Additionally, we are not assuming $P$ is true for all $k$, we are only assuming it to be true for a particular $k$. In notation, the recursive step is:

$$\forall k \in \mathbb{N} \ (P(k) \implies P(k + 1))$$

Finally, induction is not the process of proving those two statements. Those can be proved like any other statement. Induction is the statement "if those two statements are true, then $P$ is true for all natural numbers".

An example

We want to show that $\sum_{i = 1}^n i = \frac{n(n+1)}{2}$. (In terms of sets, this would be "all natural numbers are in the set $\{ n \mid \sum_{i = 1}^n i = \frac{n(n+1)}{2} \}$, but usually, the proposition syntax is easier).

"$P$ is true for $1$" is pretty easy to see, just plug it in and check.

The recursive case isn't much harder. Assume that $\sum_{i = 1}^k i = \frac{k(k + 1)}{2}$ is true for some $k$. Then, by adding $k + 1$ to both sides, we get $\sum_{i = 1}^{k+1} i = \frac{(k + 1)(k + 2)}{2}$. So, $P(k) \implies P(k+1)$.

Since $P(1)$ and $P(k) \implies P(k + 1)$, then by induction, $P(n)$ is true for all $n$.

Strong induction

The induction described above is called "weak induction". Strong induction is when your recursive hypothesis is replaced with "if $P$ is true for all $i$ less than $k$, then $P$ is true for $k$". The justification is similar, except to prove that $P(k)$ is true, we can look at any $i$ less than $k$, not just $k - 1$. It is called strong because your hypothesis is stronger, you can look at more $i$.

Sometimes this format is easier to use than weak induction. As an example, we show that all natural numbers can be written as $2^a m$, where $m$ is odd. The base case is easy, $1 = 2^0 \cdot 1$.

For the recursive case: let $k \in \mathbb{N}$ and assume that $P(i)$ is true for all $i < k$. If $k$ is odd, we are done. If not, there is some $i < k$ such that $2i = k$. By hypothesis, $i = 2^a m$ for some $a, m$, where $m$ is odd. Thus, $k = 2^{a + 1} m$.

Since $P(1)$ and $\forall i < k \ P(i) \implies P(k)$, by strong induction, $P(n)$ is true for all $n \in \mathbb{N}$. This would be difficult with weak induction.

Why is it valid?

Above, an intuitive argument for induction was shown. But in math we like to prove things. To do so, we introduce the "well-ordering principle": all non-empty subsets of $\mathbb{N}$ have a smallest element.

The well-ordering principle, weak induction, and strong induction are all equivalent!

Either you can take one of them as an axiom, or you can prove one by whichever construction of $\mathbb{N}$ you have. Then, there are pretty simple proofs to show the equivalence of the three.

Well-ordering implies weak induction: Let $S$ be a set, where $1 \in S$, and $k \in S \implies k + 1 \in S$. Let $T = \mathbb{N} \setminus S$, that is, all natural numbers not in $S$. Assume $T$ is non-empty. By well-ordering, it has a least element, $k$. Clearly it can't be $1$, because $1 \in S$. So it must be greater than $1$, and so $k - 1 \in \mathbb{N}$. Since $k$ is the smallest element in $T$, $k - 1$ can't be in $T$, meaning $k - 1 \in S$. But by assumption, that implies $(k - 1) + 1$ is in $S$, and so $k \in S$, which contradicts that $k \in T$. Therefore, $T$ was empty after all, and so $\mathbb{N} \subseteq S$.

Weak induction implies strong induction: Let $P(k)$ be a proposition where $P(1)$ and if $P(i)$ for all $i < k$, then $P(k)$. Let $Q(n)$ be the proposition "$P(k)$ is true for all $k < n$". We will use weak induction on $Q$. First, note that $Q(1)$ is true. Next, assume that $Q(n)$ is true. That means that $P(i)$ is true for all $i < k$, and so by our initial description of $P$, $P(n + 1)$ is true. But since $Q(n)$ and $P(n + 1)$ are true, $Q(n + 1)$ is true as well. Therefore, the conditions for weak induction are satisfied, and $Q(n)$ is true for all $n$. This clearly means that $P(n)$ is true for all $n$ as well.

Strong induction implies well-ordering: Let $P(n)$ be the proposition "if $n \in S$, then $S$ has a least element". Clearly it is true for $1$, because if $1 \in S$, then $1$ is the least element of $S$. For the recursive step, assume that $P(i)$ is true for all $i < k$. Then we wish to show that $P(k)$ is true. Let $k \in S$. If $k$ is the smallest element, then we are done. If not, then there is some $i < k$ such that $i \in S$. Thus, $P(i)$ is true, and so $S$ has a smallest element. By strong induction, if there is any $n \in S$, then it has a least element (this is where the "non-empty" part crops up).

Which one of these proofs you need depends on what your construction of $\mathbb{N}$ is or what axioms you have.

[Math] Tips on constructing a proof by induction.

When proving inequalities by induction it’s often very helpful simply to ask yourself what happens to each side of the proposed inequality when $n$ is increased by $1$.

In this case the expression $2n+1$ on the left increases by $2$ when you replace $n$ by $n+1$, while the expression $2^n$ on the right is multiplied by $2$ when you replace $n$ by $n+1$: instead of adding $2$, you’re adding $2^n$. If $n\ge 1$, then $2^n\ge 2$, and replacing $n$ by $n+1$ adds at least as much to the righthand side as it does to the lefthand side: it adds $2^n$ to the righthand side and $2$ to the lefthand side, and $2^n\ge 2$.

At this point you’ve already essentially done the induction step: you’ve shown that if $2n+1\le 2^n$ and $n\ge 1$, then $2(n+1)+1\le 2^{n+1}$. All that remains is to find the smallest $n\ge 1$ for which $2n+1\le 2^n$, and by actual experiment that turns out to be $n=3$.