Definitions, Conjectures, and Counterexamples

Definitions decide what belongs

A square belongs to the rectangle family. To see why, start with the meanings of the words. A quadrilateral is a closed plane figure made of four straight sides that meet only at neighboring endpoints. A rectangle is a quadrilateral with four right angles. A square is a rectangle with four congruent sides. As before, congruent segments have equal lengths.

A definition gives the exact meaning of a term. You check every condition, not how familiar the picture looks. A square with side length $3$ has four right angles and four equal side lengths. It meets the rectangle definition and the square definition. A rectangle with side lengths $2,5,2,5$ meets the rectangle definition, but its four sides are not congruent, so it is not a square.

These definitions are inclusive: a more specific family can belong to a larger family. Turning a square on the page does not remove any condition from its definition. We will not add a rule saying that a rectangle must have unequal neighboring sides.

A counterexample is an example that satisfies a claim’s conditions but fails its conclusion. The $2$-by-$5$ rectangle refutes the claim that every rectangle is a square. A triangle would not refute it: that claim is only about rectangles. One counterexample defeats an “every” claim; several successful examples do not prove one.

Observation, assumption, and proof

Suppose you examine rectangles with neighboring side lengths $2$ and $5$, $3$ and $8$, and $4$ and $6$. Each has two longer sides and two shorter sides. These examples suggest the claim that every rectangle has two longer sides and two shorter sides. Using examples to suggest a general claim is inductive reasoning. The proposed claim is a conjecture, and these examples alone do not prove it. A square with side length $3$ is also a rectangle, but all four sides have the same length, so it refutes this conjecture.

Measurements can suggest patterns too, but they have limited precision. They cannot establish exact equality by themselves.

Using accepted facts and valid reasoning to reach a conclusion is deductive reasoning. For an angle already given to be bisected, equal measures follow from the definition of angle bisector. You do not need to remeasure the drawing.

A postulate, also called an axiom, is a statement accepted as a foundation of the geometry we are studying. A theorem is a statement established by a proof from definitions, accepted foundations, and results already proved. Definitions fix meanings; postulates are accepted without proof within this course. Accepting a postulate does not make it a theorem; a theorem needs a proof.

Here is a reference to the status of ideas used so far. Keep the conditions attached to each entry.

The square-to-rectangle claim is justified here directly by the definitions: being a square already includes every rectangle condition. We have not used a theorem about diagonals, parallel lines, or triangle angles.

A conditional and its converse

Return to the claim that every square is a rectangle. We can express it as: “If a figure is a square, then it is a rectangle.”

An if-then statement, or conditional, has a condition called its hypothesis and a promised result called its conclusion. Here the hypothesis is “the figure is a square” and the conclusion is “the figure is a rectangle.” The hypothesis of a conditional is a condition to check, not an assertion that a particular figure satisfies it.

Let $p$ stand for “the figure is a square” and $q$ for “the figure is a rectangle.” The notation $p\to q$ reads “if $p$, then $q$.” It does not claim that $q$ happens only when $p$ happens.

The converse reverses the two parts: “If a figure is a rectangle, then it is a square,” or $q\to p$. Our $2$-by-$5$ rectangle satisfies its hypothesis but fails its conclusion. The original statement is true; its converse is false.

For a conditional about all allowed objects, look for a case with a true hypothesis and a false conclusion. That is the case that breaks the promise. An object outside the hypothesis does not refute it. A figure that is not a square tells us nothing against “every square is a rectangle.”

Inverse and contrapositive

The word “not” also changes a statement. Write $\neg p$ for “not $p$.” It means the entire statement $p$ is false, not just that the drawing is unusual.

The inverse negates both parts without reversing them. For the square claim it says, “If a figure is not a square, then it is not a rectangle.” This is false: the $2$-by-$5$ rectangle is not a square but is still a rectangle.

The contrapositive reverses and negates both parts. It says, “If a figure is not a rectangle, then it is not a square.” This is true: a square would have to belong to the rectangle family. A figure outside that whole family cannot belong to the square family inside it.

A conditional and its contrapositive are logically equivalent: they make the same promise. Both exclude exactly the situation ”$p$ is true and $q$ is false.” The converse and inverse likewise match each other, since each is the other’s contrapositive. Their truth need not match the original statement.

When both directions hold

A biconditional joins a conditional and its converse. The words “if and only if,” written $p\leftrightarrow q$, say that both directions hold.

For a point $M$ already on segment $\overline{AB}$, ”$M$ is the midpoint of $\overline{AB}$ if and only if $AM=MB$” is true. If $M$ is the midpoint, the lengths are equal. Conversely, if $M$ is on the segment and the lengths are equal, it meets the midpoint definition.

The location condition matters. A point away from the segment can have equal distances to the two endpoints without being their midpoint. Both the location and the equality belong in the full definition.

Valid chains of reasoning

Suppose a tile is given to be a square. Every square is a rectangle, so this tile is a rectangle. This use of a general conditional with a matching fact is the law of detachment: from $p\to q$ and $p$, conclude $q$.

If the tile is given only to be a rectangle, you cannot detach “it is a square.” That would reverse the conditional. The $2$-by-$5$ rectangle explains why the proposed step is invalid.

Now combine two true conditionals: every square is a rectangle, and every rectangle is a quadrilateral. A square meets the first condition, which leads to rectangle, which meets the next condition, which leads to quadrilateral. The law of syllogism links the chain: from $p\to q$ and $q\to r$, conclude $p\to r$. The conclusion of the first link must be the hypothesis of the next.

Common mistakes