Math 440. Modern Geometry Second Homework Assignment Fall 2010


The second assignment.

1. Exercise 1.10 (page 14).

2. Read sections 3.4 and 3.5. Then do exercises 3.1, 3.4, and 3.6.

3. Read Chapter Four and do exercises 4.1 to 4.6, 4.9, 4.13, and 4.14.


1.10 The problem in the so-called proof is that their figures are incorrect, and this leads to performing the wrong arithmetic. Here is a figure that I constructed using Geometer's Sketchpad. Note that in this figure,



the point E is outside the line segment while its counterpart point F is inside the segment . It turns out that this usually happens one point is inside the segment and the other is outside. Moreover, the point D will never be inside the triangle.

In the proof, carrying out their construction does give two congruent triangles ΔADE and ΔADF. And their reasoning showing that ΔBMD and ΔCMD are congruent is sound. Their third pair of triangles, ΔBDE and ΔCDF, are also congruent. The problem is only in their final claim:

although, and , it does not follow that .


3.1 I have a question about this one. How many of you read the assigned sections of Chapter Three before you attempted this one? My guess, based on how poorly you did on this one, is that hardly anyone did.

(a) The negation of there exists a model for incidence geometry in which the Euclidean Parallel Postulate holds

is in every model for incidence geometry the Euclidean Parallel Postulate fails.


(b) The negation of in every model for incidence geometry there are exactly seven points

is there is a model for incidence geometry in which there are not exactly seven points.


(c) The negation of every triangle has an angle sum of 180

is at least one triangle has an angle sum other than 180.


(d) The negation of it is hot and humid outside

is it is not hot or it is not humid outside.


(e) The negation of my favorite color is red or green

is my favorite color is neither red nor green.


(f) The negation of if the sun shines, then we go hiking

is the sun shines and we do not go hiking.


(g) The negation of all geometry students know how to write proofs

is some geometry student does not know how to write proofs.


3.4 (a) In the conditional "If it rains, then I get wet", the hypothesis is "it rains" and the conclusion is "I get wet".

(b) In the conditional "If the sun shines, then we go hiking and biking", the hypothesis is "the sun shines" and the conclusion is "we go hiking and biking".

(c) In the conditional "If x > 0, then there exists a y such that y2 = 0", the hypothesis is "x > 0" and the conclusion is " there exists a y such that y2 = 0".

(d) In the conditional "If 2x + 1 = 5, then either x = 2 or x = 3", the hypothesis is "2x + 1 = 5" and the conclusion is "either x = 2 or x = 3".

(e) In the conditional "Every triangle has angle sum less than or equal to 180", the hypothesis is "ΔABC is a triangle" and the conclusion is "angles ABC, ACB, and BAC add up to 180".


3.6. (a) Statement: If it rains, then I get wet.

Converse: If I get wet, then it rains.

Contrapositive: If I do not get wet, then it does not rain.

(b) Statement: If the sun shines, then we go hiking and biking.

Converse: If we go hiking and biking, then the sun shines.

Contrapositive: If we do not go hiking or do not go biking, then the sun does not shine.

(c) Statement: If x > 0, then there exists a y such that y2 = 0.

Converse: If there exists a y such that y2 = 0, then x > 0.

Contrapositive: If for every y, y2 ≠ 0, then x ≤ 0.

(d) Statement: If 2x + 1 = 5, then either x = 2 or x = 3.

Converse: If either x = 2 or x = 3, then 2x + 1 = 5.

Contrapositive: If x ≠ 2 and x ≠ 3, then 2x + 1 ≠ 5.

(e) Statement: Every triangle has angle sum less than or equal to 180.

I will first reword this in conditional form:

For every object, if the object is a triangle, then its angle sum is less than or equal to 180.

Converse: If its angle sum is less than or equal to 180, then it is a triangle.

Contrapositive: If its angle sum is greater than 180, then it is not a triangle.


4.1 For the sets A = {3, 6, 9, 12} and B = {2, 4, 6, 8, 10, 12}:

(a) A B = {2, 3, 4, 6, 8, 9, 10, 12}

(b) AB = {6, 12}

(c) = {3, 9}

(d) = {2, 4, 8, 10}


4.2. Written as a decimal, 24/7 = .


4.3. The decimal 2.357157157157 . . . written as a fraction:

Let x = 2.357157157157 . . .

Then 1000x = 2357.15715717 . . . .

Lining these two values up and subtracting gives:


1000x = 2357.157157157

x = 2.357157157

999x = 2354.8


So x = 2354.8/999 = 23548/9990.

You can reduce this fraction to 11774/4995.


4.4 There are lots of irrational numbers between 2.5834556 and 2.5834557.

Here is one: 2.583455601001000100001000001 . . . .


4.5 A rational number between π ≈ 3.141592654 and π + 0.00001 ≈ 3.141602654?

Here's one: 3.141595.


4.6 Functions f : . . .

(a) The function given by is onto but not one-to-one.

(b) The function given by f (n) = 2n is one-to-one but not onto.

(c) The function f (n) = 1 is neither one-to-one nor onto.

(d) The function f (n) = n is a one-to-one correspondence.


4.9 Given: f : (0, 1) → (1, ∞) defined by f (x) = 1/x.

One-to-one: Suppose that for some input(s) x1 and x2 we have 1/x1 = 1/x2. Then I must show that x1 = x2.

Well, from the given equality we get the equation

And so x1 = x2, QED.

Onto: Let y (1, ∞). I must show that there exists an x in the set (0, 1) for which f (x) = y.

Set x = 1/y.

Since y is positive, then x = 1/y is also positive. And since y > 1, then x = 1/y < 1. Thus, x (0, 1).

And f (x) = 1/x = 1/(1/y) = y.


4.13 The empty set has an upper bound. For example, the number one is (vacuously) greater than every element of the empty set. But this set does not have a least upper bound, in fact every number is an upper bound for this set. This might seem at first to contradict the Least Upper Bound Postulate, but that Axiom requires that the given set A be nonempty.


4.14 Let A = { x | x < }. Then this set is nonempty and has an upper bound. (For example, 2, 1.5, 1.45, and 1.4141 are all upper bounds for A.) But the least upper bound for this set is not a rational number; it's the irrational number . To be a little more formal about it, no matter what rational number r we could come up with that is an upper bound for this set, there will be another rational number between and r, and so the value r couldn't be a least upper bound.