I was at lunch with some colleagues some days ago, and we started discussing about maths. Predictably, the conversation turned to the quite specific concept of proofs used by mathematicians. I had some insights, on the reasons for so many misunderstandings between mathematicians and others. I think many of them arise from a different understanding of the word "proof"

I want to exemplify on the classical result about the cardinality of the power set, so let us dive into math.

First, the power set ${P}(A)$ of a given finite set $A$ is the the enumeration of all the possible subsets of $A$.

If $A$ is, say, the set of $\{me, you\}$ the power set will contain:

- the empty subset $\emptyset$
- the subset containing only $me$
- the subset containing only $you$
- the subset containing both $me$ and $you$

It is a legitimate and somewhat interesting question: how many elements does the power set $A$ has, if $A$ has $n$ elements? Such counting questions are very common in math and constitute the basis of combinatorics. Let us look how we can come to a solution of the problem and, even cooler, a formal proof that the solution is correct.

**Examples**
The first step is, more often than not, to list some examples. For instance, if the set $A$ only contains one elements, the power set will contain:

- the empty set
- the subset consisting of that element

We now have an example for a 1 element set (2 subsets) and for a 2 element set (4 subsets). What about a 3 element set?

By the way, we have implicitly agreed on the working assumption that the cardinality of the power set only the depends on the cardinality of the original set. This is ok, if this is wrong, we will change it. (But yes, it will turn out that this is right).

Back to our example of our 3 element set $\{me, you, the\ others\}$. We have all which we have considered before:

- the empty subset $\emptyset$
- the subset containing only $me$
- the subset containing only $you$
- the subset containing both $me$ and $you$

and in addition, all of those plus $the\ others$

- the subset containing only $the\ others$
- the subset containing only $me, the\ others$
- the subset containing only $you, the\ others$
- the subset containing both $me, you$ and $the\ othes$

This totals to 8 elements. Let us summarize: for 1 element, 2 subsets, for 2 elements, 4 subsets and for 3 elements, 8 subsets. This looks suspiciously like $2^n$, with $n$ the cardinality of $A$. This will be our working hypothesis.

**Insight**
Having formulated our working hypothesis, we should ask ourself whether this can be possibly right. Upon thinking we recognize this: for every subset in our collection, every single element of $A$ can be present or not. Since we have $n$ elements, every element, we must make $n$ binary decisions of whether to include that element in the subset or not. This gives $2 \times 2 \times ... \times 2 $ possibilities. Yes! This is $2^n$. This is our insight, and typically it is enough if you are a physicists.

But it is not a proof. It is not a proof, as to be a proof it must be formalized in a standard way using the laws of logics. This is important, since this gives all mathematicians the chance of recognizing a proof as such.

**The proof**
Let us prove this, then. The standard way to prove stuff about natural numbers is the

method of induction, which we will follow step by step.

** **First we formulate our theorem in details.

*If $A$ is a finite set with $n$ elements, then its power set $P(A)$ has $2^n$ elements.*
We prove that this is true for 0. If the set $A$ is empty, then it has 0 elements, and one single subset, which is the empty set. $1 = 2^0$. Check.

To avoid being caught in misunderstandings and quarrels about 0 and empty sets, we will prove it also for a single element. This is not required, but we do it anyway. If $A$ has a single element, its subsets are the empty set and $A$ itself. $2 = 2^1$. Check.

Now let us go to the induction step. Let us assume, we known that if $A$ has $n$ elements, then we have $2^n$ distinct subsets of $A$. Let us reason about a hypothetical set $B$ with $n+1$ elements and let us pick one arbitrary element, which we will call $b$ out of $B$. Let us remove $b$ from $B$ and obtain $B'$ a set of obviously $n$ elements. We know from the induction assumption that $B'$ has $n$ elements and thus $2^n$ subsets, constituting the set $P(B')$. Let us now consider the set $X = \{S \cup \{b\} | S \in P(B')\}$.

Interestingly, $X$ and $P(B')$ are disjoint, as every set in $X$ contains $b$ which per construction absent from every set in $P(B')$. So, the cardinality of $X \cup P(B')$ will be the cardinality of $X$ plus the cardinality of $P(B')$.

It remains to show that

- the cardinality of $X$ is $2^n$
- $X \cup P(B') = P(B)$

For 1. we should give a bijective function for $P(B')$ to $X$, which turns out to be exactly the prescription with which we constructed $X$ in the first place. This kind of clerical tasks is typically left to the reader as an exercise, and we will comply with the mathematical tradition.

For 2. we notice that if $Y$ is a subset of $B$ it either contains $b$ or not. If not, it is in $P(B')$ and thus in $X \cup P(B')$. If yes, it equals to $Y \setminus b \cup \{b\}$. Now, $Y \setminus b$ only consists of elements of $B$ apart of $b$, so it is in $P(B')$ and when we unit sets in $P(B')$ with $\{b\}$ we get by construction sets in $X$, so $Y$ is a set in $X$ and thus in $X \cup P(B')$. We just proved that $P(B) \subset X \cup P(B')$. The converse direction is left to the reader an exercise, as it follows the same principles. We also do not want to spoil him of the incredible feeling of finishing a proof.

**Conclusion**
What I hope I could convey is that sheer amount of work needed to go from an insight, giving us just a working hypothesis and some ideas, to a complete mathematical proof

**. **The real question could be here: why bother? Indeed our insight was right, and the loosely described proof ($n$ binary decisions giving $2^n$ different possibilities) is perfectly fine.

I think the problem is often that this kind of insight often fails with more complex arguments, even in

seemingly intuitive cases.

Formalized, standardized proofs in maths are a powerful method to fight the natural tendency of the human mind to fall victim of cognitive illusions and are thus one of the milestones of the human thinking!