A * conditional proof* uses assumptions which go
beyond the basic facts of mathematics. The conclusion of a
conditional proof is true *provided that* those additional
assumptions are true. Many mathematical results have the form of
"if-then" statements. A proof of such a statement will
often be a conditional proof.

* Example.* Consider the following statement:

\qquad If , then .

To prove it, you *assume* that . *Make what you've got look like what you
want:*

Likewise,

If I add and , I get . 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, , so adding
*this* inequality to , I obtain

Notice the *conditional* nature of the conclusion. It's
certainly not true that *for any*
real number x. (For example, it's false if .) The conclusion is true *if* the assumption
is true. The assumption goes beyond the basic facts of mathematics.

The last example shows how you write a conditional proof. In this
situation, you're trying to prove a statement of the form , 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 is true.

* Example.* Premises:

Prove: .

To prove the conditional statement , I *assume* the antecedent and try to deduce the consequent .

The *conclusion* was deduced on
line 8. Together with the *assumption* in line 3, this proves the conditional .

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

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

An *odd number* is an integer which can be written in the form
for some integer m. Since n is odd, for some m.

Squaring n, I get

Since is divisible by 4, leaves a remainder of 1 when it is divided by 4.

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

* Example.* If a and b are integers, a * divides* b means that there is an integer c such
that . a divides b is written for short.

For example, because , because , and since .

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

Suppose a, b, and c are integers. I'll prove that if and , then .

Suppose that and .Since , there is an integer m such that . Since , there is an integer n such that .

Substitute into to obtain , or . Since is an integer, it follows from the definition of divisibility that .

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
{\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.* You know that is always true if P is false: Anything follows from a
false assumption. For example, I'll prove:

\qquad If , then .

(Notice that both the antecedent and the consequent are false!)

Assume that . Multiply both sides by to obtain

Adding 100 to both sides, I obtain . Hence, if , then .

But note that I can do the following as well: Start with . Differentiate both sides to obtain . is true, even though 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 {\it 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.* Consider the following example from
basic algebra. I'm going to solve the equation

All the steps that follow are legitimate:

The *possible* solutions are and . You can check that works. However, if ,

So doesn't work. You may have been told that
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 , and from this
I deduced that or . In other words, this is a conditional proof of the
if-then statement

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

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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Copyright 2009 by Bruce Ikenaga