Download Jan 12 Notes

Some Building Blocks of Proof
If-Then Statements
A basic building block of mathematical statements involved in proofs are implications, or “if –
then” statements. Most theorems and axioms have this general form.
Examples:
•
If a triangle is equilateral, it is also isosceles.
•
If an increasing sequence is bounded, it has a least upper bound.
•
If a topological space is both compact and Hausdoff, then it is normal.
Letting p and q represent statements, we abbreviate “If p, then q” by p 6 q. The statements p and
q can be rather complex, and are generally statements or propositions about some mathematical
object. If p is “X is isosceles,” we might abbreviate p as p(X).
Sometimes, “if – then” statements are phrased in other ways and need to be translated:
Statement
Translation
Men are mortal.
If a man, then mortal.
Squares are rectangles.
If a square, then a rectangle.
It always rains when a cat eats grass.
If a cat eats grass, it’s going to rain.
Only those who pay tithing will get a temple
recommend.
If you have a temple recommend, you payed
your tithing.
Only those with tickets will be admitted.
If you are admitted, you had a ticket.
All lines contains exactly two points.
If there is a line, it has exactly two points.
Some Comments on “if – then” Statements:
•
p 6q is true whenever q is true, or when both p and q are false. It is only false when p is
true and q is false.
Example: I tell my daughter, “If you clean your room I will take you to McDonalds.” She
will not accuse me of lying if I take her to McDonalds even if she doesn’t clean her room.
•
In popular use, people say “you will succeed if you work hard” to mean, “you will
succeed if and only if you work hard.” These are different statements! The first means
“hard work 6 success” while the second means both “hard work 6 success” and “success
6 hard work.”
•
How do you prove an “if – then” statement? By assuming the “if” part and then arguing
that the “then” part must follow.
The argument has the following form:
p6q
Prove:
Suppose p.
{Insert here an argument that q must follow from your assumption}
Thus, q
ˆp6q
Quantifiers
•
Consider the statement, “ A square is a rectangle.” Usually, this doesn’t mean that some
square happens to be a rectangle. It probably means that all squares are rectangles. So,
•
There are implied quantifiers in front of a lot of mathematical statements. It usually
helps to make these explicit.
Examples:
Two lines intersect in a single point.
Any two lines meet in a single point.
An exterior angle is larger than the remote
interior angles.
For any triangle, exterior angles are larger
than. . .
Some Comments on Quantifiers:
•
The only two quantifiers typically used in mathematical statements are “For all” and
“There exists”.
•
In general, this means you are usually talking about either all or at least one.
The statement, “a rhombus is a rectangle” is considered false, because it usually means, “every
rhombus is a rectangle.” However, the statement, “There is a rhombus that is a rectangle” is true.
What about the statement, “Sheep are black?”
What about “Dogs orbit the earth?”
Summary
It is important to be able to understand a mathematical statement by:
•
translating it, if possible, to an “if – then” statement, and
•
deciding what the implied quantifiers are. Usually, there is an implied “for all” in front of
the statement.
An Example:
Base angles of isosceles triangles are congruent.
Statement S
If a triangle is isosceles, then it has congruent base angles.
S: If P, then Q
Given any triangle, if it is isosceles, then it has congruent
base angles.
S: For all T, if P(T), then
Q(T)
In order to prove this, you must:
•
Pick any old triangle, and
“Let ABC be a triangle. . . .
•
Assume it is isosceles
“And suppose AB=AC”
(Or, pick any old isosceles triangle)
•
Show its base angles are congruent.
{Insert hard part of proof
here}
. . .So, pB –pC
General Example:
For Every
, if that
has Property P, then it also has Property Q.
For every X, if P(X), then Q(X)
œX(P(X) 6 Q(X))
To prove this, you must:
1.
2.
3.
Consider an arbitrary X.
Assume it has property P.
Show it has Property Q.
Example:
Prove that the diagonals of a parallelogram are congruent.
Translation: For any quadrilateral, if it is a parallelogram, then its diagonals are congruent.
You must:
1.
Pick a general quadrilateral and
2.
Assume it is a parallelogram (Property P)
“Let ABCD be a
quadrilateral. . .”
“and assume is
to
parallel
.
(Or, pick a general parallelogram)
3.
Show that its diagonals are congruent (Property Q).
{Insert hard part of proof
here}
“So, AD = BC.”
Another General Example:
For any
, if that
has property P, then there is a
with
property Q.
For any X, if X has property P, then there is a Y with property Q.
œX(P(X) 6›Y(Q(Y)))
To show this, you must:
1.
Consider an arbitrary X , and
2.
Assume it has property P, then
3.
Find (somehow) a Y and
4.
Show it has property Q.
Example:
Given pAVB and point P in its interior, show that the ray
intersects the segment
.
Translation: For any angle and point, if the point is interior, then there is a (different) point on
both
and
.
To show this you must:
1.
Consider an arbitrary angle and name it
appropriately;
“Let pAVB be an angle,”
2.
Consider an arbitrary point that
“and let X be a point that is. .
.”
3.
Is in the interior of the angle (Property P)
“interior to pAVB.”
4.
Then show that there must be a point (Y) such that
{Insert hard part of proof
here} “Thus, there is a point
Y”
5.
It lies on both the ray and the segment (Property Q).
“that lies on both the ray and
segment.”
Some Basic Proof Forms:
‚
Direct
‚
Contradiction
‚
Contrapositive
‚
By cases
We’ve shown examples of Direct proofs above. Here we discuss the general form of the other s.
Sample form of a proof by contradiction:
Given:
Prove:
p, s, etc.
q
(Note: This is equivalent to “If p, s, etc., then q”)
Suppose ~q
{Insert here arguments that arrive at a contradiction to a given, e.g. ~p, or ~s, or (t
and ~t), etc.}
ˆq
Sample of a proof by cases:
Given: p, t, w, etc.
Prove: q
(Note: This is equivalent to “If p, t, w, etc., then q”)
Either r, s, or q.
s 6 ~p
r 6 ~t
ˆq
Another sample of a proof by cases:
Prove: q
Either p or ~p
p6q
~p 6 q
ˆq
Sample proof using contrapositive:
Given: s
Prove: p 6 q
Suppose ~q.
{Insert here arguments that arrive somehow at ~p, e.g.:
(s and ~q) 6 ~p
So, ~p}
Thus, ~q 6 ~p
ˆp6q
Some Final Notes
•
To prove “p if and only if q” (p iff q), prove separately p 6 q and q 6 p.
•
To prove “There exists an x” satisfying a certain property P(x), [(› x)P(x)], you can either
find the x and show it works, or try a proof by contradiction.
•
To prove “For all x, P(x)”, [(œx) P(x)], pick an arbitrary y and show P(y). Or try
contradiction.
•
~ [(› x)P(x)] = (œx) ~P(x)
•
~ [(œx) P(x)] = (› x) ~P(x)
•
To prove that one set is the same as another, (A = B) show that each is a subset of
another: AfB and BfA.