Download Assignment 4 — LibEd3100 1. What or who were the muses after

Assignment 4 — LibEd3100
1. What or who were the muses after which Alexandria’s Museum was named?
What art was each associated with? Was one associated with mathematics?
From Wikipedia, the muses were the goddesses who inspired the creation of
literature and the arts. Variations are possible, but the generally accepted nine
are:
Muses
Calliope
Clio
Erato
Euterpe
Melpomene
Polyhymnia
Terpsichore
Thalia
Urania
Art
Epic poetry
History
Love poetry
Song and elegiac poetry
Tragedy
Hymns
Dance
Comedy
Astronomy.
There was no specific muse of mathematics, although if we take the more classical view of mathematics, then astronomy is a mathematic, in which case Urania
would be an appropriate muse. A similarly broad view would accord any of the
muses associated with song and dance a mathematics link. Polyhymnia by
some sources is also identified with geometry, making a link with mathematics
possible with this muse.
2. Prove Euclid’s Proposition I-17, that the sum of any two angles in a triangle is
less than two right angles. Using the notation in the diagram below, prove that
the sum of the interior angles B and C is less than two right angles.
Solution: We can extend the line BC out beyond C to D. Now consider the
exterior angle 6 ACD. By the Exterior Angle Theorem, it is greater than 6 ABC.
Therefore we have 6 ABC + 6 ACB < 6 ACD + 6 ACB = two right angles, so
the sum of the two inside angles at B and C is less than two right angles.
3. In class we saw how the Greeks used a diagram and geometric argument to verify
the identity (a + b)2 = a2 + 2ab + b2 . Give a similar diagram and argument for
the identity (a − b)2 = a2 − 2ab + b2 . (Hint: Start with a large square with sides
of length a, and divide the top into lengths b and a − b.)
Solution: We start with a large square with sides of length a, and divide the
top into lengths b and a − b. Divide bottom in the same way, and then divide
both sides into lengths b and a − b. Now draw in 4 rectangles, inside the large
square.
Now we can write an equation that describes the area of the square. Overall it
is a2 ; but it is also the sum of the areas of the four smaller rectangles.
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This gives a2 = b2 + 2b(a − b) + (a − b)2 . Rearrange to solve for (a − b)2 :
(a − b)2 = a2 − b2 − 2b(a − b) = a2 − b2 − 2ab + 2b2 = a2 − 2ab + b2 .
4. Prove Euclid’s Proposition I-28: If two lines intersected by a transversal have
the sum of the two interior angles on the same side of the transversal equal to
two right angles, then the two lines are parallel. (Hint: Use Prop. I-27.)
Solution: The angles α and γ sum to two right angles, and the angles α and β
are given as summing to two right angles. Since α is common to each sum, we
have α + γ = α + β so that γ = β upon subtracting α from each side.
Thus γ and β are two congruent alternate interior angles. Therefore, by Proposition I-27, the two lines intersected by the transversal t are parallel.
5. Briefly, what was the main goal of education in classical Athenian society?
Classical Athenian society used education to instill discipline in its boys, which
would be of benefit when waging war.
6. a) Use the Euclidean Division Algorithm to find the gcd of 119 and 272.
b) Use the Euclidean Division Algorithm to find the gcd of 258 and 964.
Solution: a) Following our method, we find
272 = 2(119) + 34,
119 = 3(34) + 17,
34 = 2(17) + 0.
Therefore the gcd is 17.
b) Similarly,
258
190
68
54
14
12
=
=
=
=
=
=
1(190) + 68,
2(68) + 54,
1(54) + 14,
3(14) + 12,
1(12) + 2,
6(2) + 0.
Therefore the gcd is 2.
7. Let n be any odd whole number. Prove that n2 has the form 8k + 1 for some
whole number k. (Hint: Use the Division Theorem on n and 4, and do cases
for what remainders are possible.)
Solution: Let n be any odd integer. Considering remainders on division by 4,
we know that n can be written as n = 4q + r for some quotient q and remainder
r, where r is one of 0, 1, 2 or 3. But n is odd, so it cannot be of the form 4q + 0
or 4q + 2. Therefore n must have the form 4q + 1 or 4q + 3, for some integer q.
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Now consider these two cases:
Case 1: If n = 4q + 1, then n2 = (4q + 1)2 = 16q 2 + 8q + 1, which has the form
8k + 1 for some k = 22q 2 + q.
Case 2: If n = 4q + 3, then n2 = (4q + 3)2 = 16q 2 + 8q + 9 = 8(2q 2 + q + 1) + 1
= 8k + 1 for k = 2q 2 + q + 1.
8. Use Euclid’s method to prove that there must be a prime larger than 19.
Solution: Consider the number x = (2 × 3 × 5 × 7 × 11 × 13 × 17 × 19) + 1.
This number is larger than any of the primes from 2 to 19, by construction; and
it must have a prime divisor (possibly itself). Also by construction, this prime
divisor cannot be any of the primes from 2 to 19, so it must be a prime number
larger than 19.
If you form the product, you’ll find 2 × 3 × · · · × 19 + 1 = 9699691 which factors
as 347 × 27953.
For the smaller product 2 × 3 × · · · × 17 + 1 = 510511 we have the prime
factorisation 19 × 97 × 277.
9. Find the prime factorisation of the integers 2013 and 36000.
Solution: 1000 divides into 36000, so using this fact to reduce the numbers
involved, and noting that 10 = 2 × 5, we obtain
2013 = 3 × 11 × 61
36000 = 1000 × 36 = 1000 × 9 × 4 = 23 × 53 × 92 × 22 = 25 × 32 × 53 .
Note that 61 is prime.
10. Show that the only prime p for which 3p + 1 is a perfect square is p = 5. (Hint:
If 3p + 1 = a2 , then 3p = a2 − 1 = (a + 1)(a − 1).)
Solution: Following the hint, 3p = (a − 1)(a + 1) for some integer a, so 3p is
represented as a product of two integers. The only factorisations of 3p are as
1 × 3p,
or 3 × p,
or p × 3,
or 3p × 1,
so need only consider the possibilities for matching up these with a − 1 and
a + 1.
In the first instance, a−1 = 1 and a+1 = 3p implies a = 2 and so 3p = a+1 = 3
from which p = 1, which is impossible since then p would not be prime.
In the second instance, a − 1 = 3 and a + 1 = p from which a = 4 and so p = 5.
In the third instance, a − 1 = p and a + 1 = 3 so a = 2 and a − 1 = 1 = p, again
an impossibility if p is to be prime.
Finally, for a − 1 = 3p and a + 1 = 1 we obtain a = 0 giving 3p = −1, another
impossibility.
Thus, only p = 5 is possible, corresponding to a = 4.
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