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Assignment 3 — LibEd3100
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in Roman numerals.
1. How did Romans denote fractions? Write 9 12
Romans generally wrote out fractions in natural language, so 59 would be expressed as “five ninths.” However, most routine fractions were in twelfths, and
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for this dots and an “S” were used to represent the fractions: one dot for 12
,
2
5
6
two dots for 12 and so on up to 12 . For 12 , an “S” was used, and then more
11
7
on up to 12
.
dots added after the S to represent 12
7
So, 9 12
would be represented as IX S·
2. Thales was regarded as one of the seven sages of Greece. Who were the others,
and what were they noted for? (Be brief.)
Quoting Wikipedia, the other six sages were:
• Cleobulus of Lindos: “Moderation is the best thing.” He governed as tyrant
of Lindos, in the Greek island of Rhodes, circa 600 BC.
• Solon of Athens: “Keep everything with moderation.” Solon (640-559 BC)
was a famous legislator and reformer from Athens, framing the laws shaped
the Athenian democracy.
• Chilon of Sparta: “You should not desire the impossible.” Chilon was a Spartan politician from the 6th century BC, to whom the militarization of
Spartan society was attributed[citation needed].
• Bias of Priene: “Most men are bad.” Bias was a politician and legislator in
the 6th century BC.
• Pittacus of Mytilene (c. 650 BC), governed Mytilene (Lesbos) along with
Myrsilus. He tried to reduce the power of the nobility and was able to
govern with the support of the popular classes, whom he favoured. He
famously said “You should know which opportunities to choose.”
• Periander of Corinth: he was the tyrant of Corinth in the 7th and 6th centuries
BC. Under his rule, Corinth knew a golden age of unprecedented stability.
He was known saying “Be farsighted with everything.”
3. The Pythagoreans knew how to make a Pythagorean triple starting with any
number n you want. For instance, for n an odd positive whole number, you can
use the triple (n, (n2 − 1)/2, (n2 + 1)/2).
a) What triple does this produce for the number n = 7?
b) Why do we need to have n odd for this triple to work? (What happens if
you try n = 4?)
c) Verify that the triple (n, (n2 −1)/2, (n2 +1)/2) is indeed always a Pythagorean
triple.
Solution: (a) For n = 7, this gives the triple (7, 24, 25).
(b) We need n odd here in order for (n + 1)/2 and (n − 1)/2 to be natural
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numbers.
(c) Verify that the triple (n, (n2 −1)/2, (n2 +1)/2) is indeed always a Pythagorean
triple:
2
n +
µ
n2 − 1
2
¶2
1
1
= n2 + (n4 − 2n2 + 1) = (4n2 + n4 − 2n2 + 1)
4
4
µ 2
¶2
1 2
n +1
1 4
2
2
(n + 2n + 1) = (n + 1) =
,
=
4
4
2
as required.
4. Verify that the number 1225 is both a triangular and a square number. (Triangular numbers are described in the coursepack on p. 119.)
Solution: First, 1225 = 352 , so 1225 is a square number, in fact, the 35th
square number. To see that it is also a triangular number, we need to find a
natural number n such that tn = 1225. Using tn = n(n + 1)/2, we need to solve
n(n + 1)/2 = 1225.
Therefore n(n + 1) = 2450, which can be rearranged to n2 + n − 2450 = 0.
Using the quadratic formula gives us
p
−1 ± 1 + 4(2450)
−1 ± 99
n=
=
,
2
2
giving two possible answers of 98/2 = 49 or −50. Since we need a positive
number here, our solution is 49.
Therefore 1225 is both the 35th square number and the 49th triangular number.
5. a) Prove that if a triangular number tn is multiplied by 8 and 1 is added, the
result is a square number. (This result was stated by Plutarch in about 100
AD.)
b) Prove that if tn is a triangular number, then 9tn + 1 is also a triangular
number.
Solution: (a) Let n be an integer, and let tn = n(n + 1)/2.
Then 8tn + 1 = 8(n)(n + 1)/2 = (8n2 + 8n + 2)/2 = 4n2 + 4n + 1 = (2n + 1)2 ,
which is the 2n + 1th square number. So 8tn + 1 = s2n+1 .
(b) Let tn = n(n + 1)/2 be the nth triangular number.
Then 9tn + 1 = 9(n)(n + 1)/2 + 1 = (9n2 + 9n + 2)/2 = (3n + 1)(3n + 2)/2.
This last number is the triangular number t3n+1 !
√
6. Argue that 3 is irrational.
√
Solution: Following the pattern used in the lecture for demonstrating that 2
is irrational, we begin
Suppose, to
√ by constructing a proof by contradiction.
√
the contrary, that 3 is rational, so that we can write 3 = m/n for two whole
numbers m and n. Without loss of generality, we can assume that m and n
2
share no common factors, as these can be cancelled without changing the value
of the ratio.
√
Squaring 3 = m/n produces 3 = m2 /n2 so 3n2 = m2 . 3 divides the left-hand
side of this, so that 3 must then divide the right-hand side, m2 , and so 3 divides
m. Thus m = 3k for some whole number k and so 3n2 = m2 = (3k)2 = 9k 2 .
Dividing through by 3 produces n2 = 3k 2 and again, we see that since 3 divides
the right-hand side, 3 must also divide the left-hand side n2 , so 3 divides n, too.
Therefore there must be a whole number p for which n = 3p.
This tells us m = 3k and n = 3p for some whole numbers k and p, which
contradicts the assumption that m and n had no common factors (apart from
1, of course).
√
√
Thus, we cannot write 3 = m/n for whole numbers m and n, i.e., 3 is
irrational.
Alternatively, from 3n2 = m2 , with m and n in lowest terms, we observe that
not both m and n can be even (for otherwise we can repeatedly divide out a 2
until at least one of m or n is odd). If one of m or n is even, then the other
must be also, so we need only consider the possiblity that m and n are both
odd.
Thus, we may write m = 2k + 1 and n = 2p + 1 for some whole numbers k and
p, so 3n2 = m2 implies
3(2p + 1)2 = (2k + 1)2
or 3(4p2 + 4p + 1) = 4k 2 + 4k + 1
or
12p3 + 12p + 3 = 4k 2 + 4k + 1.
Subtracting 1 from each side gives
12p2 +12p+2 = 4k 2 +4k
and dividing by 2 produces 6p2 +6p+1 = 2k 2 +2k.
In this last equation, for any whole numbers p and k, the left-hand side is always
odd, while the right-hand side is always even, so the equation is impossible.
√
∴ 3 6= m
for any whole numbers m and n.
n
7. Theodorus of Cyrene lived in about 400 BC and taught mathematics to Plato.
For any natural number n he gave two methods, depending
on whether n is
√
odd or even, to construct a line segment of length n. For any odd natural
number n, the method is to draw a right-angled triangle with hypotenuse of
length (n + 1)/2 and one side of length (n − 1)/2.
a) Draw a picture of this situation.
√
b) Verify algebraically the claim that the remaining side then has length n.
Solution: Consider the triangle ABC in which angle B is a right angle, and
side AB has length (n − 1)/2 while the hypotenuse AC has length (n + 1)/2.
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Note that since n is odd, both these side lengths are natural numbers. Then
the third side has a length x which must satisfy
µ
¶2
¶2
µ
n−1
n+1
2
+x =
.
2
2
Therefore
¶2 µ
¶2
µ
1
1
n−1
n+1
2
−
= (n2 + 2n + 1 − n2 + 2n − 1) = · 4n = n.
x =
2
2
4
4
√
So the third side has length x where x2 = n, in other words, has length n.
8. What does the English word “peripatetic” mean? What language does it come
from, and what is its connection to mathematics and science?
Solution: The English word “peripatetic” means “Walking about or from place
to place; traveling on foot” or a person who walks about, an itinerant (from
Answers.com). Mathematically, it refers to a person who follows the philosophy
of Aristotle, who walked about while teaching his lessons in the Lyceum. Word
origin: Greek peripatetikos, from peripatein, to walk about, or from peripatos,
covered walkway, which was where Aristotle allegedly lectured.
9. a) What argument form does the following argument use?
“If I am taking History of Math, then I must be interested in History. I am
taking History of Math. Therefore I am interested in History.”
(Give the name and the symbolic pattern using letters.)
b) Write out the argument form for the Disjunctive Syllogism, and give an
example in words of its use.
Solution: (a) This is the Modus Ponens argument form, symbolized as “If P
then Q, and P, therefore Q”. (See the last section of Ch. 5 of the lecture notes.)
(b) The Disjunctive Syllogism form is “P or Q, but not P, therefore Q”.
For an example, consider: “My child is a boy or a girl. It is not a boy. Therefore
it is a girl.”
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