1.3 Reasoning and Logic
It was in geometry that the human race discovered that it could reason in very precise, logical ways. This discovery naturally led to an extensive analysis of what constituted valid reasoning and what constituted fallacious reasoning. The point was to establish ways by which one could start with a true statement and reason with confidence to a true statement. Let $TS$ stand for a true statement, $FS$ stand for a false statement, and let the arrow stand for valid reasoning. Then it must be the case that $TS\rightarrow TS$, and it must be impossible for $TS\rightarrow FS$, since otherwise we could never trust our reason. What are the other possibilities — could we have $FS\rightarrow FS$ and $FS\rightarrow TS$? Consider the following examples:
$TS\rightarrow
TS$ |
$FS\rightarrow
FS$ |
$FS\rightarrow
TS$ |
\begin{align*} 1&=1\\ 1+3&=1+3\\ 4&=4 \end{align*} |
\begin{align*} 1&=2\\ 1+3&=2+3\\ 4&=5 \end{align*} |
\begin{align*} 1&=2\\ 2&=1\\ 1+2&=2+1\\ 3&=3 \end{align*} |
We see that clearly, both $FS\rightarrow FS$ and $FS\rightarrow TS$ are possible; but while we may have expected to see $FS\rightarrow FS$, the third example ($FS\rightarrow TS$) is somewhat surprising. In the case of $FS\rightarrow TS$, we reasoned this way: Although $1=2$ is false, it is nonetheless true that if $a=b$ then $b=a$, and on that basis we obtain the second step. Then we add equals to both sides and arrived at a true result. So one can reason validly from a false statement to a false statement or from a false statement to a true statement.
It is a pity that many students fail to understand two major implications of the above analysis. The first major implication is this: Let $P$ be a statement whose validity we are trying to prove, i.e., we don't know if $P$ is true. Suppose we reason from $P$ to a true statement; that is, we want to show $P\rightarrow TS$. Does that mean that $P$ is true? The answer, to be blunt, is not necessarily. The reason? Both $TS\rightarrow TS$ and $FS\rightarrow TS$ are possible. Here's an example of a `proof' that fails to account for the above implication.
Prove that for $a,\;b>0$, $\dfrac{a+b}{2}\leq\sqrt{ab}$.
$$
\begin{align*}
\dfrac{a+b}{2}&\geq\sqrt{ab} && P\\
(a+b)^2&\geq\left(2\sqrt{ab}\right)^2 && \text{Mulitply by 2 and square two positive numbers.}\\
a^2+2ab+b^2&\geq4ab && \text{Expand.}\\
a^2-2ab+b^2&\geq0 && \text{Subtract $4ab$ from both sides.}\\
(a-b)^2&\geq0 && \text{True Statement.}\\
\end{align*}
$$
The last statement is true since any real number squared is greater than or equal to 0. But this is not a proof since $P$ could still be true or false. However, not all is lost; in fact, much has been gained. If we can reverse our reasoning, if we can start with our true conclusion and reason backwards to $P$, then we can, in fact, conclude that $P$ is true. But we have to be able to reverse our reasoning. This means that we can treat reasoning from $P$ to $TS$ as a way to discover a valid proof, and that, surely, is quite useful. Here's a correct proof.
$$
\begin{align*}
(a-b)^2&\geq0 && \text{True statement.}\\
a^2-2ab+b^2&\geq0 && \text{Expand or mulitply out.}\\
a^2+2ab+b^2&\geq4ab && \text{Add $4ab$ to both sides.}\\
(a+b)^2&\geq\left(2\sqrt{ab}\right)^2 && \text{Factor and express as a square.}\\
\frac{a+b}{2}&\geq\sqrt{ab} && \text{Take the square root and divide by 2.}\\
\end{align*}
$$
The second major implication to be drawn from the fact that we have $TS\rightarrow TS$, $FS\rightarrow FS$, and $FS\rightarrow TS$, but not $TS\rightarrow FS$. The only way we can reason correctly from P and obtain a false conclusion is if $P$ itself is false. This means that if we start with a statement $P$ whose truth we don't know and reason until we reach a false conclusion, then we can conclude that $P$ itself is false. This is the basis for indirect proof, a topic we'll take up in chapter 3.
Deductive reasoning is essentially the stringing together of if-then statements, each of which is justified. It looks like this:
If $A$ is true, then $B$ is true
because . . .
Since $B$ is true, then $C$ is true because . . .
Because $C$ is true, then $D$ is true because . . .
$D$ implies $E$ because . . .
Since $B$ is true, then $C$ is true because . . .
Because $C$ is true, then $D$ is true because . . .
$D$ implies $E$ because . . .
Reasons are expressed in if-then statements as well. Basically, a statement such as "If $A$ is true, then $B$ is true" is a specific instance of some reason, and since the reason is valid in all instances, it justifies this particular instance.
Let us turn our attention to if-then statements. Let $P$ stand for one statement and $Q$ stand for another. Let $\sim P$ stand for the opposite or negation of $P$ and $\sim Q$ stand for the opposite or negation of $Q$. Thus, if $P$ stands for "$x=3$", then $\sim P$ stands for "$x\neq3$." It should be clear that that we are working with a two-valued logic; i.e., if $P$ is true, then $\sim P$ is false and vice versa. In addition, we will be working with statements in which universality is claimed. That is, we will be claiming that a statement is true in all cases. This means that if the statement is true for some cases, but false for others, then the statement itself is false. There is no 'maybe true' in this geometry. Either a proposition is always true or it is false. This means that in order to show that a statement is false, it is sufficient to find one counterexample. Often a diagram will do the job. For example, to show that "if a figure is a rectangle, then its diagonals are perpendicular" is false, it is sufficient to draw a rectangle in which the diagonals are obviously not perpendicular. But one or two examples do not and cannot prove an assertion. This can't be over-emphasized.
- Definition(If... , then...): In the statement, "If $P$, then $Q$," $P$ is called the hypothesis and $Q$ is called the conclusion. The statement may also be expressed as "$P$ implies $Q$" which is symbolized by $P\rightarrow Q$.
There are three important if-then statements.
Statement:
If $P$, then $Q$.
Converse: If $Q$, then $P$.
Contrapositive: If $\sim Q$, then $\sim P$.
Converse: If $Q$, then $P$.
Contrapositive: If $\sim Q$, then $\sim P$.
Consider these examples:
Statement: If $B$ is the midpoint of $\overline{AB}$, then $AB=BC$.
Converse: If $AB=BC$, then $B$ is the midpoint of $\overline{AC}$.
Contrapositive: If $AB\neq BC$, then $B$ is not the midpoint of $\overline{AC}$.
The three diagrams below suggest that while the statement and its contrapositive are both true, the converse is false since we've shown an example where it is not true.
Statement: If two sides of a triangle are equal, then the triangle is equilateral.
Converse: If a triangle is equilateral, then two sides are equal.
Contrapositive: If a triangle is not equilateral, then no two sides are equal.
The statement is false since the triangle could be isosceles but not equilateral. The converse is true since if all three sides are equal, then certainly two sides are equal. The contrapositive is false since the triangle could still be isosceles.
Statement If $x=3$, then $4x=12$.
Converse If $4x=12$, then $x=3$.
Contrapositive If $4x\neq12$, then $x\neq3$.
Clearly, all three are true.
We can formulate some conjectures from these examples. The first is that a statement and its contrapositive seem to be either both true or both false. The second is that a statement and its converse may both be true or one may be true and the other false.
These conjectures are, in fact, the case and we can see why if we express if-then statements in terms of Venn diagrams. Visually, think of an if $P$, then $Q$ statement as asserting that if $x$ lies in set $P$, then it lies in set $Q$. The statement if $\sim Q$, then $\sim P$ asserts that if $x$ does not lie in $Q$ then it does not lie in $P$.
The leftmost diagram illustrates that $P\rightarrow Q$ is true since if $x$ lies in $P$, it clearly lies in $Q$. Now consider the converse: $Q\rightarrow P$. It would be true if the left-hand diagram were the case, but it would be false if the middle diagram were the case. Since it could be either true or false, we must declare it false. The right-hand diagram represents the contrapositive, that is that if $x$ doesn't lie in $Q$ then it doesn't lie in $P$. Note that if the statement is true and we have sets $P$ and $Q$ related as in the left-hand diagram, then the contrapositive must also be true.
Thus, we can conclude that a statement and its contrapositive are logically equivalent, that is they are either both true or both false. This implies that we can substitute one for the other. This is useful. It is sometimes the case that a statement is either difficult to understand or difficult to prove, while the contrapositive is easier on both counts. We also note that a true statement may have a false converse. The importance of this can't be overemphasized. Students often start giving reasons without using the if-then form. They often will give as a reason 'alternate interior angles' without specifying whether they mean 'if alternate interior angles are congruent, then the lines are parallel' or 'if two lines are parallel, then the alternate interior angles are congruent'. That sloppiness not only leads them to misunderstand the meaning and application of some theorems, but can easily lead them to formulate false ideas — not the best approach to mastery of the material.
Sometimes a statement and its converse are both true. A Venn diagram of that instance would simply show that the circles representing the elements of $P$ and $Q$ were exactly the same. Whenever "If $P$, then $Q$"and "If $Q$, then $P$" are both true, we can combine them into a single statement of the if and only if form, namely "P if and only if ." This is often abbreviated with iff, as in "$P$ iff $Q$." For now let's assume that the following statements are both true:
If a triangle is equilateral, then it is equiangular.
If a triangle is equiangular, then it is equilateral.
They can be combined as: A triangle is equilateral if and only if it is equiangular. Note that an iff statement does not begin with the word "if."