# When Writing Your Own Proofs...

## CS390, Fall 2019

**Abstract**

My experience is that many students are uncomfortable with proofs, but we are going to be seeing a lot of proofs in this course and you will be responsible for writing a number of them.

Here I discuss some common mistakes and problems that I have encountered in student-written proofs.

# 1 A Proof is Mathematics PLUS Words

In general a proof is an argument to convince the reader of a conclusion. Its purpose is to convince the reader. It’s not the reader’s responsibility to sift through the argument for pieces that can be re-arranged to make sense.

I often see student-written proofs that consist of line after line of mathematical equalities, with no obvious relation from one to the next (and no English words to explain how they were supposed to be related).

In many of these cases, I truly believed that they *were* simply unrelated lists of “facts” (e.g., formulas copied from the textbook), after which appeared a simple assertion that, altogether, these proved the theorem. In other cases, I *suspected* that there was some coherent idea in the student’s mind, but could only guess at what it might be.

That doesn’t work. You have to

- link your steps together logically, and you have to
*communicate*those logical links to me.

That communication generally requires adding text between the steps of mathematics.

# 2 Respect the Constraints

I may often ask for a proof using a specific technique.

In that case, I expect you to give me a proof using **that** method, even if the theorem is already proven in the textbook using a different style of proof.

# 3 You Can’t Prove (Most) Theorems By Giving an Example

Consider the Theorem: *For all natural numbers $k$, if $k$ is odd then $k^2$ is odd.*

You *cannot* prove this theorem by giving an example of a particular value of $k$, e.g.,

Let $k=3$.

$3$ is odd.

$k^2 = 9$, and $9$ is odd. QED (not!)

This “proof” tells us only about one particular value of $k$, but the theorem says that we want to make its claim **“for all”** natural numbers. That wording of “for all” or “for every” tells us that “proof by example dies not work”.

To really prove this theorem, we need to use statements that are true for all possible values of $k$, e.g.:

If $k$ is odd, then $\exists m \in \cal{N}$ (there exists a natural number $m$) such that $k = 2m+1$.

Then $k^2 = (2m+1)^2 = 4m^2 + 4m + 1$. Now let $n = 2m^2 + 2m$. Then $k^2 = 2n+1$, and $k$ must be odd. QED

Consider, as another example, the theorem: *For all natural numbers $k$, $k+1 = 2k$.*

We can certainly find an example where this is true: let $k=1$.

But that does not prove the theorem is true for all $k$. In fact, if we had chosen *any* other example for $k$, the statement $k+1 = 2k$ would not have worked out. This illustrates my next point:

## 3.1 You Can *Disprove* “for all” Theorems by Example

We can use examples to disprove theorems that make claims about all members of some set:

Theorem:

For all natural numbers $k$, $k+1 = 2k$.Let $k=2$. Now, $k+1 = 3$ and $2k = 4$, but $3 \neq 4$, so the theorem does not hold and is disproven.

## 3.2 You Can Prove “there exists” Theorems by Example

Although most theorems we will encounter try to state a property that holds for all elements of a large set, sometimes we will encounter existence theorems, for example:

Theorem:

There exists some natural number $k$ for which $k+1 = 2k$.Let $k=1$. Now, $k+1 = 2$ and $2k = 2$, so $k+1 = 2k$. QED

The difference between $\forall$ (for all) and $\exists$ (there exists) is crucial.

# 4 Proving Equalities

Many proofs involve showing two expressions to be equal. You’ve been doing that since you first studied algebra, but there are some subtleties to chaining together a series of steps that actually make sense.

You can’t prove that $\mbox{expression}_1$ equals $\mbox{expression}_2$ by writing

\[ \begin{align} \mbox{expression}_1 = & \mbox{expression}_2 \\ \mbox{expression}_{1a} = & \mbox{expression}_{2a} \\ \mbox{expression}_{1b} = & \mbox{expression}_{2b} \\ \vdots & \\ \mbox{expression}_{n} = & \mbox{expression}_{n} \\ \end{align} \]

in which the a, b, … variants are applied by applying the same operation to both the left-hand and right-hand sides of the preceding line, until the left-hand and right-hand sides have been reduced to something identical on both sides.

A proof like this results in a series of perfectly correct steps that add up to *absolutely nothing*.

This does **not** prove that $\mbox{expression}_1$ equals $\mbox{expression}_2$. Instead, it proves that **if** $\mbox{expression}_1$ equals $\mbox{expression}_2$, then $\mbox{expression}_n$ equals $\mbox{expression}_n$. But that’s kind of useless, since we already know that any expression equals itself.

For example, suppose that I wished to show, as part of a proof by induction, that, if $\sum_{i=0}^{k} 1 = k + 1$ then it must be true that $\left(\sum_{i=0}^{k+1} 1\right) = k + 2$.

Here is an **invalid** proof in a form that I see all too often:

Prove that

\[ \left(\sum_{i=0}^{k+1} 1\right) = k + 2 \]

Expand the sum:

\[ \left(\sum_{i=0}^{k} 1\right) + 1 = k + 2 \]

We have stipulated that $\sum_{i=0}^{k} 1 = k + 1$, so substituting

\[ \left( k + 1 \right) + 1 = k + 2 \]

Subtract $k+1$ from both sides.

\[ 1 = 1 \]

which is, obviously true.

Again, this is an invalid proof. To show that is it invalid, let me use the same technique to “prove” that 2 equals 1.

Prove:

\[ 2 = 1 \]

The symmetric property of equality gives

\[ 1 = 2 \]

And, because I can add equal things to equal things,

\[ 2 + 1 = 1 + 2 \]

Subtracting 2 from each side,

\[ 1 = 1 \] which is obviously true. QED, we have shown that $2 = 1$.

The problem with these “proofs” is that they start by asserting the very equality that we want to prove. If that assertion is, in false, we can still prove all manner of true (and false) conclusions from it.

How should you structure such a proof? Try:

\[ \begin{align} \mbox{expression}_1 & = \mbox{expression}_{1a} \\ & = \mbox{expression}_{1b} \\ & = \mbox{expression}_{1c} \\ & \vdots \\ & = \mbox{expression}_{2} \\ \end{align} \]

where each right-hand side is derived from the expression before it by some obviously true manipulation.

For example:

Prove that, if $\sum_{i=0}^{k} 1 = k + 1$ then it must be true that $\left(\sum_{i=0}^{k+1} 1\right) = k + 2$.

Expanding the sum:

\[ \left(\sum_{i=0}^{k+1} 1\right) = \left(\sum_{i=0}^{k} 1\right) + 1 \]

We have stipulated that $\sum_{i=0}^{k} 1 = k + 1$, so

\[ \begin{align} \left(\sum_{i=0}^{k+1} 1\right) & = (k + 1) + 1 \\ & = k + 2 \\ \end{align} \]

QED

Notice that the left-hand side of the equalities we wrote never changed.

# 5 What’s Wrong with These?

Have “proven”, above, that $2 = 1$, you might be interested in some of these other botched proofs. See if you can spot the flaws.

## 5.1 More Proofs that 2 = 1

Still more ways to prove that $2 = 1$:

### 5.1.1 Proof 1

\[ 2 = 1 \]

Multiply both sides by zero:

\[ 0 = 0 \]

This is obviously true, so the theorem is proven.

### 5.1.2 Proof 2 (a classic)

Let $a$ be any number and let $b = a$.

\[ a = b \]

Multiply both sides by $a$.

\[ a^2 = ab \]

Subtract $b^2$ from both sides:

\[ a^2 -b^2 = ab - b^2 \]

Factor each side:

\[ (a + b)(a - b) = b(a - b) \]

Cancel (divide by) the common factor $b-a$:

\[ (a + b) = b \]

Now, let $a = b = 1$,

\[ (1 + 1) = 1 \]

\[ 2 = 1 \]

## 5.2 Prove that all horses are brown.

*Theorem: All horses are the same color.*

We’ll prove this by induction:

**Base case**: Take any set containing one horse. Clearly all horses in that set are the same color.

**Inductive case** Assume that any set of $k$ horses are the same color. Prove that any set of $k+1$ horses must be the same color.

Consider the set of $k+1$ horses ${h_1, h_2, \ldots, h_{k+1}}$.

Now, remove the first horse and consider the set ${h_2, \ldots, h_{k+1}}$. This has $k$ horses and, by the inductive hypothesis, all are the same color.

Returning to the original set, remove the last horse and consider the set ${h_1, h_2, \ldots, h_{k}}$. Again, this set has $k$ horses and, by the inductive hypothesis, all of them are the same color.

Now, $h_1$ is the color as ${h_2, \ldots h_k}$ and $h_{k+1}$ is the color as ${h_2, \ldots h_k}$. So all of ${h_1, h_2, \ldots h_k, h_{k+1}}$ are the same color and the inductive case is proven.

Q.E.D.

*Corollary: All horses are brown.*

All horses are the same color. Secretariat was brown. Therefore all horses are brown.

## 5.3 Prove that Julius Caesar had six limbs.

Shakespeare has recorded that, on his way to the Forum, Julius Caesar was accosted by a soothsayer. “Beware the Ides of March!”, the date on which a group of Roman senators would assassinate him.

Caesar was therefore forewarned of the assassination attempt.

By proverbial wisdom, “Forewarned is forearmed”.

Therefore Julius Caesar had 4 arms. Add to that his two legs, and he had a total of six limbs.