Conditional Proofs

A conditional proof is a proof of an "if-then" (conditional) statement.

Since any proof makes some assumptions, you might say that every proof is a conditional proof. But there are different kinds of assumptions you might make. Consider the following conditional statements:

"If $x \in \real$ , then $x^2 + 49 \ge 14 x$ ."

"If $x \in \real$ and $x > 3$ , then $x^2 +
   5 x + 2 > 25$ ."

There's an obvious sense in which the "if" part of the second statement has more content that the "if" part of the first statement. In the first statement, the "if" part doesn't give you much to go on. In the second statement, you would suspect that the number 7 is somehow related to the "then" part.

In this section, I'll look at proofs of conditional statements where the "if" part carries that kind of significant information.


Example. Prove that if $x \in \real $ and $x > 3$ , then $x^2 + 5 x + 2 > 25$ .

To prove the statement, you assume that $x > 3$ . Now

$$x > 3, \quad\hbox{so}\quad x^2 > 9.$$

Likewise,

$$x > 3, \quad\hbox{so}\quad 5 x > 15.$$

If I add $x^2 > 9$ and $5 x > 15$ , I get $x^2 + 5 x > 24$ . I compare this inequality to the target inequality, and I see that I'm missing a "2" on the left and a "1" on the right. Well, $2 > 1$ , so adding this inequality to $x^2 + 5 x > 24$ , I obtain

$$x^2 + 5 x + 2 > 25.$$

Notice the conditional nature of the conclusion. It's certainly not true that $x^2
   + 5 x + 2 > 25$ for any real number x. (For example, it's false if $x = 0$ .) The conclusion is true if the assumption $x > 3$ is true.


The last example shows how you write a conditional proof. In this situation, you're trying to prove a statement of the form $P \ifthen Q$ , where P is the set of assumptions --- it may be one statement, or several statements --- and Q is the conclusion. You assume P and try to derive Q. If you succeed, then $P \ifthen Q$ is true.


Example. Premises: $\left\{\matrix{A \land \lnot D
   \cr B \ifthen (C \ifthen D) \cr}\right.$

Prove: $(A \ifthen B)
   \ifthen\, \lnot C$ .

To prove the conditional statement $(A \ifthen B) \ifthen\,\lnot C$ , I assume the "if" part $A \ifthen B$ and try to prove the "then" part $\lnot C$ .

$$\matrix{ \hfill 1. & A \land \lnot D \hfill & \hbox{Premise} \hfill \cr \hfill 2. & B \ifthen (C \ifthen D) \hfill & \hbox{Premise} \hfill \cr \hfill 3. & A \ifthen B \hfill & \hbox{Premise for conditional proof} \hfill \cr \hfill 4. & A \hfill & \hbox{Decomposing a conjunction (1)} \hfill \cr \hfill 5. & B \hfill & \hbox{Modus ponens (1,3)} \hfill \cr \hfill 6. & C \ifthen D \hfill & \hbox{Modus ponens (2,5)} \hfill \cr \hfill 7. & \lnot D \hfill & \hbox{Decomposing a conjunction (1)} \hfill \cr \hfill 8. & \lnot C \hfill & \hbox{Modus tollens (6,7)} \hfill \cr \hfill 9. & (A \ifthen B)\ifthen\, \lnot C \hfill & \hbox{Conditional proof (3,8)} \hfill \cr}$$

The conclusion $\lnot C$ was deduced on line 8. Together with the assumption $A \ifthen B$ in line 3, this proves the conditional $(A \ifthen B) \ifthen\,\lnot C$ .


Example. Prove that if n is odd, then $n^2$ leaves a remainder of 1 when it is divided by 4.

I assume that n is an odd number. I want to prove that $n^2$ leaves a remainder of 1 when it is divided by 4.

An odd number is an integer which can be written in the form $2 m + 1$ for some integer m. Since n is odd, $n = 2 m + 1$ for some m.

Squaring n, I get

$$n^2 = (2 m + 1)^2 = 4 m^2 + 4 m + 1 = 4(m^2 + m) + 1.$$

Since $4(m^2 + m)$ is divisible by 4, $n^2$ leaves a remainder of 1 when it is divided by 4.

This proves the conditional statement that if n is odd, then $n^2$ leaves a remainder of 1 when it is divided by 4.


Example. ( Proving the contrapositive) Let n be an integer. Prove that if $3 n^2 + 5 n + 18$ is not even, then n is not even.

Recall that the conditional $P \ifthen Q$ is logically equivalent to its contrapositive $\lnot Q \ifthen \lnot P$ . In some cases, you use this to prove a conditional statement by replacing it with its contrapositive.

In this example, the given conditional statement is kind of awkward: Both the "if" and "then" parts are negative statements. And if I try to prove this conditional statement by assuming the "if" part --- that is, assuming that $3 n^2 + 5 n + 18$ is not even --- it isn't obvious how to proceed.

Instead, I replace the statement with its contrapositive: "If n is even, then $3 n^2 + 5 n +
   18$ is even."

Begin by assuming the "if" part: Suppose n is even. By definition, this means that $n = 2 m$ , where m is an integer. Then

$$\eqalign{ 3 n^2 + 5 n + 18 & = 3 (2 m)^2 + 5 (2 m) + 18 \cr & = 12 m^2 + 10 m + 18 \cr & = 2(6 m^2 + 5 m + 9) \cr}$$

Now $6 m^2 + 5 m + 9$ is an integer, so $2(6 m^2 + 5 m + 9)$ is even (by definition of "even"). Hence, $3
   n^2 + 5 n + 18$ is even.

Since I've proved the contrapositive, this proves the original statement.


If a and b are integers, a divides b means that there is an integer c such that $ac = b$ . a divides b is written $a \mid b$ for short.

For example, $6 \mid 18$ because $6\cdot 3 = 18$ , $10 \mid 0$ because $10 \cdot 0 = 0$ , and $-6 \mid 6$ since $(-6) \cdot (-1) = 6$ .

On the other hand, 3 does not divide 5, since there is no integer c such that $3 \cdot c = 5$ . "3 does not divide 5" is written $3 \notdiv 5$ .

Example. Suppose a, b, and c are integers. Prove that if $a \mid b$ and $b \mid c$ , then $a
   \mid c$ .

Suppose that $a \mid b$ and $b \mid c$ .Since $a
   \mid b$ , there is an integer m such that $a m = b$ . Since $b \mid c$ , there is an integer n such that $b n =
   c$ .

Substitute $a m = b$ into $b n = c$ to obtain $(a
   m) n = c$ , or $a (m n) = c$ . Since $m
   n$ is an integer, it follows from the definition of divisibility that $a \mid c$ .

Note that when I introduced m and n, I was careful to use different letters than the a, b, and c that were already in use. Note also that after stating my assumptions, I translated those assumptions using the definition of divisibility. I asked myself: "What does it mean for a to divide b? What does it mean for b to divide c?"

When a proof involves a concept (like divisibility in this proof), you must have a clear idea of what it means. You should be able to write down a clear, correct definition before proceeding; if you can't remember the definition, look it up in your book or notes and write it down. Just having a vague idea of what something means is not enough when you're writing proofs.


Example. Prove: "If $0 = 1$ , then $\pi =
   100$ ."

Notice that both the "if" and "then" parts are false!

Assume that $0 = 1$ . Multiply both sides by $\pi - 100$ , then do some algebra:

$$\eqalign{ 0 \cdot (\pi - 100) & = 1 \cdot (\pi - 100) \cr 0 & = \pi - 100 \cr \pi & = 100 \cr}$$

Hence, if $0 = 1$ , then $\pi = 100$ .

Note that I can do the following as well: Start with $0 = 1$ . Differentiate both sides to obtain $0 = 0$ . $0 = 0$ is true, even though $0 = 1$ is false.

People sometimes erroneously suppose that they can prove something by starting with what they want to prove, and working until they get a true statement --- which they suppose provides "confirmation" that the original statement is true. This example shows that the procedure is invalid, since a true conclusion may come from a false assumption.

Assuming what you want to prove is the most fundamental logical fallacy. It's called begging the question. In a formal debate, the question was the issue or statement being debated. Thus, begging the question referred to arguing for the truth of the statement by assuming that it was true. It's also called circular reasoning. Another way to see that this procedure is nonsensical is to observe that if you're trying to prove that P is true and you assume that P is true, then you're done: There's no need for an argument at all!


Example. Solve the equation in $\real$ :

$$x + 3 = \sqrt{x + 5}.$$

All the steps that follow are legitimate:

$$\eqalign{ (x + 3)^2 & = (\sqrt{x + 5})^2 \cr x^2 + 6 x + 9 & = x + 5 \cr x^2 + 5 x + 4 & = 0 \cr (x + 1)(x + 4) & = 0 \cr}$$

The possible solutions are $x = -1$ and $x = -4$ . You can check that $x = -1$ works. However, if $x = -4$ ,

$$x + 3 = -1, \quad\hbox{while}\quad \sqrt{x + 5} = 1.$$

So $x = -4$ doesn't work. You may have been told that $x = -4$ is an "extraneous solution". But if you think about this, you might wonder how it is that a sequence of valid algebraic steps can lead to an incorrect solution.

To understand this, think about what I just proved. I assumed that $x + 3
   = \sqrt{x + 5}$ , and from this I deduced that $x = -1$ or $x = -4$ . In other words, this is a conditional proof of the if-then statement

$$\hbox{``If $x + 3 = \sqrt{x + 5}$, then $x = -1$ or $x = -4$.''}$$

However, you know that a conditional statement $P \ifthen Q$ and its converse $Q \ifthen P$ aren't logically equivalent. Therefore, there is no reason that the following statement should be true:

$$\hbox{``If $x = -1$ or $x = -4$, then $x + 3 = \sqrt{x + 5}$.''}$$

And in fact, as I noted above, it's false!

In other words, when you "solve an equation" in algebra, you're really finding a set of solution candidates. This will not be exactly the set of solutions unless the steps in your derivation are reversible.

One of the steps in the derivation above is not reversible, and that is what leads to the "extraneous solution". Can you tell which step it is?


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