Download MATH 1200 Section B — Prof. Madras Problem for Tutorial and

MATH 1200 Section B — Prof. Madras
Problem for Tutorial and Homework — March 21, 2016
INSTRUCTIONS: Problem 18.A(i,ii,iii ) should be worked on during the tutorial. Submit your solution at the conclusion (either on paper or by sending a
photo to [email protected]). Problems 18.A(iv,v ) and 18.B are for
homework, due by noon on Friday April 1. The homework problems may be
done individually or in a group of up to 4 students. All the students in the
group will get the same grade for the work. You can submit the homework to
me personally, or by e-mail, or to my mailbox in N520 Ross.
18.A. The question of the day is: which positive rational numbers can be written
as a sum of reciprocal of distinct positive integers? For example, 2/3 and 6/7
can each be written in this form, as follows:
1
1
2
=
+
3
2
6
6
1
1
1
=
+ +
.
7
2
3
42
and
We shall restrict our attention to rational numbers between 0 and 1. AccordA
where A and B be natural numbers
ingly, consider a rational number x = B
with 0 < A < B. Is the following true:
(**) The rational number
A
B
can be written
A
1
1
1
=
+
+ ··· +
B
n1
n2
nk
where n1 , n2 , . . . , nk are k different natural numbers, for some
k ≥ 1.
(Remark : We insist that n1 , . . . , nk are all different because otherwise the question is too easy: e.g. 32 = 13 + 13 .)
A
Here is an algorithm that tries to find such an expression for a given B
with
0 < A < B.
• Let x1 = A/B. Let n11 be the largest rational number of the form
1
that is less than or equal to x1 (i.e., satisfying n11 ≤ x1 < n11−1 .)
n
• Let x2 = x1 − n11 . If x2 6= 0, let n12 be the largest rational number
of the form n1 that is less than or equal to x2 .
• Repeat until xi = 0.
1
(It is not obvious that this algorithm will ever stop; maybe the xi ’s will get
smaller and smaller without ever being 0. We will investigate this below.) You
can check that this algorithm works for 23 and 76 , giving the sums shown earlier.
(i) What does the algorithm produce for
A
B
=
5
13
?
The assertion (**) is obviously true when A = 1: simply take k = 1. In the
A
tutorial, we shall consider the case A = 2 (and B odd, so that B
is in lowest
terms.)
(ii ) Assume A = 2 and B is an odd integer greater than 2. Show that the above
algorithm always works, giving a representation with two terms:
1
1
2
=
+
.
B
n1
n2
(iii ) Show that the representation in part (ii ) is not always unique. That is, for
some odd B > 2, it is possible to write B2 = m11 + m12 , where the integers m1
and m2 are different from the ones that the algorithm gives you. (Each group
should find a different example.)
........................................
The parts below are for homework:
(iv ) In the algorithm, show that x2 =
0 ≤ α < A.
α
β
where α and β are integers with
(v ) Use part (iv ) to prove that the algorithm must eventually stop, and that
A
less than 1.
the statement (**) is true for every positive rational number B
A
18.B. Let x = B
be a positive rational
number in lowest terms (i.e., A, B ∈ N
√
and hcf(A, B) = 1). Prove that x is rational if and only if A and B are both
perfect squares.
(Suggestion: Use Propositions 11.3 and 11.4. Remember that your proof should
work for cases like A = 40, where A is not a perfect square but has a factor
that is a perfect square.)
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*** If you have not already done so: complete the Online Tutorial on Academic
Honesty, and send the confirmation page to Prof. Madras (either by hard copy
or e-mail). ****
ADDITIONAL PROBLEM TO THINK ABOUT (not to be handed in):
18.C. Assume that x is a real number that satisfies the equation
xm + am−1 xm−1 + · · · + a1 x + a0 = 0
where the coefficients a0 , a1 , · · · , am−1 are all integers. (Note that the coefficient
of xm is 1.) Prove that x is either an integer or an irrational number.
(You may have to deal with two cases in your proof: whether a0 is 0 or nonzero.)
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