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) A ∩
B = {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 : ℕ
→ ℕ . . .
(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 ∈ ℚ | 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.