Summative workshop
Name: ______________________
If a collection of points that determines only integer distances is infinite, then the points must all
be in a line.
The proof will be by contradiction so let’s explore “contradiction” a bit. A logical contradiction
is an assertion that a statement and its negation are simultaneously true.
Examples:
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Currently it is raining in Olympia and it is not raining in Olympia.
The defendant was at the scene of the crime and the defendant was not at the scene of the
crime.
Two is an even number and two is an odd number.
x belongs to a set and x does not belong to a set.
the number x is a rational number and x is the square root of two.
the set S is finite and S is not finite (infinite).
Note that, “ All triangles are potatoes and some sound waves smell green,” while nonsensical
and false, is not a contradiction. A contradiction must be of the form “is / is not.” Contradictions
are resolved by identifying which of the two statements is true.
1.)
Make up an example of a mathematical and of a non-mathematical contradiction.
If the contradiction cannot be resolved, it is called a paradox. Some historical examples of
paradoxes:
a.
b.
c.
2.)
Everything I say is a lie. (The Cretan paradox.)
Autological means that a word describes itself. For example: short, English, and
polysyllabic are autological. Heterological means that a word does not describe
itself. For example: loud, monosyllabic, and living are heterological. Into
whichcategory does heterological fit? (Russell’s barber paradox.)
Ignore this sentence. (Hofstader).
Analyze the paradoxical nature of each of the three examples.
Let S be a set that determines only integer distances. This is given and guaranteed. We need to
prove the following:
if S is infinite, then the elements of S must be in a line.
Proof by contradiction works by setting up two mutually exclusive outcomes, showing that one
outcome leads to a contradiction (making it the wrong choice of outcomes), and declaring the
other outcome true, by elimination.
The points of S are either aligned or not aligned. Let’s try “not aligned” and see what that leads
to.
S is an infinite set that determines only integer distances and the points of S are not aligned.
Then there must be three points A, B, and C that form a non-degenerate triangle.
3.)
Sketch.
Any other point of the set cannot be a vertex of  ABC and hence cannot be on two sides of the
triangle. (Only vertices are simultaneously on two sides.) So if P is any other point in S, it fails to
be on two sides of the triangle, say AB and AC.
4.)
Explain from the hypotheses, why A, B, and C must exist and why the other point P is
not just made up.
What are the consequences of P not being on AB? Add point P to your sketch.
Since P ≠ C,  ABC ≠  PAB. What do we know about triangles? The triangle
inequality!
| PA – PB | < AB, which is by hypothesis some integer. Since all distances determined by S are
integers,
| PA – PB | = AB – 1, or
| PA – PB | = AB – 2, or
| PA – PB | = AB – 3, or …
| PA – PB | = 2, or
| PA – PB | = 1.
Absolute value is related to distance. For example, | PA – PB | = 3 means: the distance from P to
A minus the distance from P to B equals three. This means that P is on an hyperbola with major
axis 3 and foci at A and B.
Taken together, any point of S distinct from A, B, and C and not on AB must be on one of the
AB – 1 hyperbolas described above.
But , any point of S distinct from A, B, and C must also fail to be on another side of
 ABC (other than AB) – we have designated this other side AC.
5.)
Repeat the above argument for P not on AC
Since P is not on AB and not on AC, it must be on one of the hyperbolas
| PA – PB | < AB and one of the hyperbolas you described in 5.), that is, it is on the
intersection of two hyperbolas.
Each condition generated finitely many hyperbolas;
finitely many hyperbolas intersect at finitely many points; and
any point P in S but distinct from A, B, and C must be one of the hyperbolic intersection points.
Therefore, …
Hence, the contradiction …
Hence, “not aligned” is the wrong choice.
And the conclusion is …
6.)
Fill in the above blanks.