# Mid1Soln.pdf

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University of Toronto Scarborough

Mathematics

MATA31H3

Leo Goldmakher

Fall

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U NIVERSITY OF TORONTO SCARBOROUGH
MATA31H3 : Calculus for Mathematical Sciences I
MIDTERM EXAMINATION # 1
October 3, 2012
Duration – 2 hours
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SIGNATURE: Midterm Exam # 1 MATA31H3 page 2 of 8
(1) (20 points) Use induction to prove that
Xn
4 = 4 (4 ▯ 1) 8n 2 N:
3
k=1
Xn
k 1 2 n
[Recall that4 is a fancy way of writing 4 + 4 + ▯▯▯ + 4 .]
k=1
Proof: By induction.
Let n n o
X k 4 n
A = n 2 N : 4 = (4 ▯ 1) :
k=1 3
4
1 2 A, since 4 3 (4 ▯ 1).
Now, suppose n 2 A. Then from the deﬁnition of A,
Xn
k 4 n
4 = 3 (4 ▯ 1):
k=1
n+1
Adding 4 to both sides yields
n+1
X k 4 n n+1
4 = (4 ▯ 1) + 4
k=1 3
n+1 n+1
4 ▯ 4 3 ▯ 4
= +
3 3
4 ▯ 41▯ 4
=
3
4n+2▯ 4
=
3
4 n+1
= (4 ▯ 1):
3
Thus, we see that whenever n 2 A, we must have n + 1 2 A as well. By induction,
we conclude that A = N, which is precisely what is claimed. ▯
continued on page 3 Midterm Exam # 1 MATA31H3 page 3 of 8
p
(2) (20 points) Prove tha5 62 Q.
There are (at least!) two different approaches:
p
Proof 1: Suppose 5 2 Q. Then 9a 2 Z and 9b 2 N such that a is fully reduced,
b
and p a
5 = :
b
Squaring both sides and simplifying shows that
2 2
(y) a = 5b :
2
In particular, a is a multiple of 5. It follows (see Lemma below) that a is a multiple
of 5. So, we can write a = 5k for some k 2 Z. Plugging this back into (y) and
simplifying yields
2 2
b = 5k :
Exactly as above, this implies that b is a multiple of 5. Thus, both a and b must be
a
multiples of 5. But this contradicts our assumptionbtis fully reduced! ▯
2
Lemma: Suppose a 2 Z, and a is a multiple of 5. Then a is also a multiple of 5.
Proof. Let k be the smallest integer such that 5k > a. (Such an integer exists by the
Well-Ordering Property of N.) This implies that 5(k ▯ 1) ▯ a, since otherwise, k
wouldn’t be minimal. So, we have
5k ▯ 5 ▯ a < 5k:
This means that
a = 5k ▯ r;
where r = 1;2;3;4; or 5. Moving things around and squaring both sides, we see
that
2 2 2
r = a ▯ 10kr + 25k :
Note that the entire right hand side is a multiple of 5, since a is a multiple of 5 by
2 2
hypothesis, and 10kr and 25k are clearly multiples of 5. We thus conclude that r
must be a multiple of 5. There are only ﬁve possible values of r, and it’s easy to
check that the only one of these for which r is a multiple of 5 is r = 5. Thus, we
conclude that a = 5k ▯ 5, which is a multiple of 5 as claimed. ▯
—————————-
—————————-
p
Proof 2: Let A = fn 2 N : n 5 2 Zg. If A = ;, then we’re done.
Suppose instead that A 6= ;. By the Well-Ordering Property of N, A must have
some least element; call it a. I now claim that
p
(*) a( 5 ▯ 2) 2 A:
p
Since a( 5 ▯ 2) < a, this would contradict the minimality of a.
So, why is (*) true? First, note that a( 5 ▯ 2) 2 Z, since a 5 2 Z and 2a 2 Z. Next,
p
since a( 5 ▯ 2) > 0, we deduce that it is actually a natural number. Finally, we
have
p p p
a( 5

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