Download Selected Solutions to Homework 5 1. Complete the worksheet

Selected Solutions to Homework 5
1. Complete the worksheet handed out in class on Wednesday. Be sure to
write up the construction steps and proofs that they create the desired
information carefully.
Comment: For the most part the solutions on this were good. A
couple of brief comments are warranted however. Be sure to at know
how the constructions work. In particular, don’t just say copy the
angle, be sure that you explain how to do so. When you are teaching,
this is part of what you will need to do.
2. In class last week√
we defined the geometric mean of two numbers a and
b to be equal to ab. Show that if f (x) = ln(x), then the geometric
mean of the numbers a and b is the value c such that f (c) is the average
of the values of f (a) and f (b). Would this still be true if we used the
common logarithm (base 10) in place of the natural logarithm? Explain
why or why not.
√
Solution:
Let
√
√ c = ab. Then f (a) = ln(a), f (b) = ln(b), and f (c) =
f ( ab) = ln( ab). Now by the power rule for natural logs,
√
1
f (c) = ln( ab) = ln (ab) 2
1
=
ln(ab)
2
1
=
(ln(a) + ln(b)),
by the product rule for logs
2
1
=
(f (a) + f (b))
2
Thus f (c) is the average of f (a) and f (b).
If you replace the natural logarithm with any logarithm (whether the
logarithm base 10 or any other) then because the only properties of
the natural logarithm we used in the proof hold for all logarithms, the
result would be true for any logarithm.
3. Suppose you have invested $50 in an account with continually compounding interest. The banker tells you that according to his table,
after 5 years you will have exactly $125. After leaving the bank, you
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realize that you needed to know how much money you would have after
2 and a half years. How much money would you have if you withdraw
it then? (assume no penalties) Explain.
Solution: One method for solving this is to use the general formula
for the principal of an investment after n years at a given interest rate
P (n) = P0 ern where r is the interest rate. So that 125 = 50e5r . Then
we have r = 51 ln(5/2). Using this we can solve for P (2.5) to get the
result.
A second method would be to avoid finding the interest rate would be
to simply use that
P (2.5) = 50e2.5r
= 50(e5r )1/2 .
We now bring the 50 inside the square root to give
P (2.5) = (50 · 50e5r )1/2 = (50 · 125)1/2 .
Thus the value P (2.5) is the geometric mean of 50 and 125.
A third method would be to look at the underlying principle behind
interest calculations. Namely, the key fact about interest is that it
always accrues by the same power over equal amounts of time. That
is, if you have P0 dollars at the start, and after time t you have P0 · a
dollars, then at time 2t you have P0 · a2 . Consequently, using t = 2.5,
and the information given in the problem
q the amount of money after
2
, and the amount 50a =
5 = 2t years is 125 = 50a . Hence a = 125
50
q
√
50 125/50 = 125 · 50 or the geometric mean of 50 and 125.
This is a general notion of the geometric mean. When a problem involves exponentiation or the property that a constant ration occurs over
a constant time, then the inverse function corresponds to a logarithm
and you have that the geometric mean arises as the half-way point.
4. Suppose you are standing at the origin, and at every lattice point (point
in the plane with integer coordinates) there is a fencepost (of 0 thickness). Is there any direction that you can look so that there is not a
fencepost in front of you? That is, if you look infinitely far, must you
see a fencepost in your direct line of sight? Explain.
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Solution: Suppose you see a fencepost at the lattice point (a, b), where
a and b are integers. Since you are standing at the point (0, 0), you are
looking along a line of slope b/a. Thus, you are looking along a line
of rational slope. Hence, if you look along a line with irrational slope,
then you will√never see a fence post. Consequently, if you look along
the line y = 2 · x, you will never see a fencepost. (In fact, if you look
in a random direction, the probability that you will see a fencepost is
0 because the rational numbers are insignificant next to the irrational
numbers.)
5. Prove that if a line has two points with rational coordinates (all four
of which are different), then any point (x, y) on the line with one coordinate rational must have the other coordinate rational.
Solution: Let the points (a, b) and (c, d) be two points on the given
line. Since these points both have rational coordinates, the slope of the
d−b
line, m = c−a
is a rational number (not equal to 0 as b 6= d and finite
since c 6= a). Now suppose (x, y) is another point on the line with x or
y−b
y rational. Then x−a
= m, so that y = m(x − a) + b. Consequently, if x
is rational, since m, a, and b are all rational and the rational numbers
are closed under multiplication, subtraction, and addition, it follows
that y is also rational. A similar argument shows that if y is a rational
number then x is a rational number too.
6. A number field is a non-empty set of real numbers that is closed under addition, multiplication, subtraction, and division by non-zero elements. In particular, 0 and 1 must be in the field. Prove that any
number field must contain all of the rational numbers. That is, show
that if F is a number field, then for every pair of integers a and b with
b 6= 0, a/b ∈ F .
Solution: Since F is a number field, 1 ∈ F . Let a be an integers. Since
|a| = 1 + 1 + . . . + 1 (a 1s) and F is closed under addition, |a| ∈ F .
If a = |a|, then a ∈ F . Otherwise, a = 0 − |a| ∈ F (since F is closed
under subtraction) and again a ∈ F . Thus every integer is in F . Let q
be a rational number. Then there exist integers a and b with b 6= 0 such
that q = a/b. Since F is closed under division by non-zero elements, it
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follows that q ∈ F . As q was an arbitrary rational number, it follows
that F contains all the rational numbers. QED
7. Show that if you can construct segments of length a and b, and you
have a pair of coordinate axes given, then you can construct the point
on the Cartesian plane with coordinates (a, b). (This is not hard, but
requires a little explanation.)
Suppose a and b are constructible lengths. Then given a pair of coordinate axes, to construct the point (a, b), do the following. First, mark
off the length a along the x-axis with your compass point at the origin.
This is then the point (a, 0). Next construct a line perpendicular to the
x axis at (a, 0) by constructing a circle centered at (a, 0) intersecting
the x-axis at points (0, 0) and (2a, 0). Using a compass length longer
than a, then draw circles centered at (0, 0) and (2a, 0) of equal radius.
These circles intersect in two points. The line connecting these two
points is the perpendicular to the x-axis at (a, 0). Now mark off the
length b on this perpendicular line with the compass point at (a, 0).
The point at this intersection has coordinates (a, b).
8. The three explanations of why a negative times a negative is a positive.
Comments: Each of the three explanations has their own best setting.
They also all have their own background. The first two explanations
begin with the notion that we have a choice of defining what a negative
times a negative should be. This is certainly one perspective. People
chose to manipulate the symbols in this way because it has, in the
first explanation, a geometric meaning, and in the second explanation,
a natural extension of a pattern with the natural numbers. Neither
explanation corresponds to a model of why a negative multiplied by
a negative should be a positive, and one of the difficulties with the
explanation is that no good elementary models exist (at least that I
know of). The third explanation is inherently different from the first
two because it starts with the assumption that we want the distributive
property for multiplication over addition to hold. The third explanation
is then a proof that if we require the distributive law to hold true then
we have no other choice but to define a negative times a negative to be
a positive.
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In terms of prior knowledge, the first explanation requires an understanding of the number line (which may or may not be the case with
the students that you will teach), an understanding of function at some
elementary level, and finally an understanding that a positive times a
negative is a negative. The last is at least easy to convince people of.
It also requires a willingness to believe that multiplication shouldn’t
collapse the number line (unless multiplying by 0). The second explanation requires less prior knowledge. It requires some experience
with patterning and arithmetic series patterns (though not under that
name). It also requires knowing that a positive times a negative is a
negative. The third explanation requires the most background, which
is probably an underlying reason why most of you chose not to use it.
In particular, a student would need some familiarity with proof, some
understanding of the distributive law on a more abstract level (even
though I avoided using −a), and a mathematical patience, by which I
mean a willingness to work through the problem without seeing how
the answer will arise. This last is an important willingness to cultivate
in students, because as you move to a higher and higher level of mathematics you will need to have more and more patience to wait for the
punchline of a proof.
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