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University of Toronto Scarborough

Mathematics

MATA31H3

Leo Goldmakher

Fall

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Carefully tear this page off the rest of your exam.
U NIVERSITY OF TORONTO S CARBOROUGH
MATA31H3 F : Calculus for Mathematical Sciences I
R EFERENCE S HEET
Properties of real numbers
Throughout this course, we assume that there exists a set R endowed with two binary operations + and ▯
satisfying all the following properties:
Algebraic Properties:
(A1) 8a;b 2 R, we have a + b = b + a.
(A2) 8a;b;c 2 R, we have a + (b + c) = (a + b) + c.
(A3) There exists a number 0 2 R such that 8a 2 R, a + 0 = a.
(A4) For each a 2 R there exists a number ▯a 2 R such that a + (▯a) = 0:
(M1) 8a;b 2 R, we have a ▯ b = b ▯ a.
(M2) 8a;b;c 2 R, we have a ▯ (b ▯ c) = (a ▯ b) ▯ c.
(M3) There exists a number 1 2 R such that 8a 2 R, a ▯ 1 = a. Moreover, 1 6= 0.
(M4) For each a 2 Rnf0g, there exists a numbe2 R such that a ▯ = 1:
(D) 8a;b;c 2 R, we have a ▯ (b + c) = a ▯ b + a ▯ c.
Order Properties: There exists a special subset P ▯ R such that
(O1) 0 62 P
(O2) For all a 2 Rnf0g, a 62 P iff ▯a 2 P.
(O3) If a;b 2 P, then a + b 2 P and a ▯ b 2 P.
Completeness Property
(C) If A ▯ R is nonempty and bounded above, then there exists a least upper bound of A.
N OTATION
supA, the supremum of A, is the least upper bound of A (if it exists).
inf A, the inﬁmum of A, is the greatest lower bound of A (if it exists).
The symbol a ▯ b means a + (▯b). The symboleans a ▯ b.
n+1 n 1
For all n 2 N, we deﬁne the symbol := a ▯ a, where a := a.
Given a;b 2 R, the notation a > b means that a + (▯b) 2 P, and b < a is another way of writing a > b. The
symbol a ▯ b means that a + (▯b) 2 P [ f0g, and b ▯ a means the same as a ▯ b. U NIVERSITY OF TORONTO SCARBOROUGH
MATA31H3 : Calculus for Mathematical Sciences I
MIDTERM EXAMINATION # 2
November 9, 2012
Duration – 2 hours
Aids: none
NAME (PRINT):
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STUDENT NO: KEY
TUTORIAL:
Tutorial section # Name of TA
Qn. # Value Score
COVER PAGE 5
1 20
2 20
3 12
4 18
5 15
6 10
Total 100
TOTAL:
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SIGNATURE: Midterm Exam # 2 MATA31H3 page 2 of 8
(1) Prove each of the following, explicitly justifying every step of your proof with the appro-
priate property of R.
(a) (8 points) 2 ▯ 2 = 4.
We have
2 ▯ 2 = 2 ▯ (1 + 1) by deﬁnition of 2
= 2 ▯ 1 + 2 ▯ 1 by (D)
= 2 + 2 by (M3)
= (1 + 1) + (1 + 1)
= 1 + 1 + 1 + 1 by (A2)
= 4 by deﬁnition of 4:
(b) (12 points) If a 2 (0;2), then a < 4.
If a 2 (0;2), then a > 0 and a < 2, or in other words,
a 2 P and 2 ▯ a 2 P:
By (O3), we deduce that a + a 2 P. Note that
(2 ▯ a) + (a + a) = 2 + (▯a) + a + a by (A2)
= 2 + 0 + a by (A4)
= 2 + a by (A3):
Since 2 ▯ a 2 P and a + a 2 P, (O3) implies that 2 + a 2 P. Applying (O3)
yet again to the two numbers 2 ▯ a and 2 + a, we deduce that
(*) (2 ▯ a) ▯ (2 + a) 2 P:
Finally, we have
▯ ▯ ▯ ▯
(2 ▯ a) ▯ (2 + a) =(2 ▯ a) ▯ 2 + (2 ▯ a) ▯ a by (D)
▯ ▯ ▯ ▯
= 2 ▯ (2 ▯ a) + a ▯ (2 ▯ a) by (M1)
▯ ▯ ▯ ▯
= 2 ▯ 2 + 2 ▯ (▯a) + a ▯ 2 + a ▯ (▯a) by (D)
= 2 ▯ 2 + 2 ▯ (▯a) + a ▯ 2 + a ▯ (▯aby (A2)
= 4 + 2 ▯ (▯a) + a ▯ 2 + a ▯ (▯a) by part (a)
= 4 + 2 ▯ (▯a) + 2 ▯ a + a ▯ (▯a) by (M1)
▯ ▯
= 4 + 2 ▯ (▯a) + a + a ▯ (▯a) by (D)
▯ ▯
= 4 + 2 ▯ a + (▯a) + a ▯ (▯a) by (A1)
= 4 + 2 ▯ 0 + a ▯ (▯a) by (A4)
= 4 + 0 + a ▯ (▯a) by proof in lecture
▯ 2 2 ▯
= 4 + a + (▯a ) + a ▯ (▯a) by (A4)
▯ ▯
= 4 + (▯a ) + a 2 + a ▯ (▯a) by (A1)
= 4 ▯ a + a + a ▯ (▯a) by (A2)
▯ ▯
2
= 4 ▯ a + a ▯ a + (▯a) by (D)
2
= 4 ▯ a + a ▯ 0 by (A4)
2
= 4 ▯ a + 0 by proof from lecture
= 4 ▯ a2 by (A3):
Combining this with (*) above, we see that 4▯a 2 P, or equivalently, a < 4.
▯
continued on page 3 Midterm Exam # 2 MATA31H3 page 3 of 8
p
(2) (20 points) Prove that 5 exists. In other words, prove that there exists a positive number
2
x 2 R satisfying x = 5. [You do not need to explicitly use the algebraic and order proper-
ties in this problem, or in any of the following problems.]
Let A := fx > 0 : x < 5g. Clearly A 6= ; (since 2 2 A). Also, observe that A is
bounded above by 3, since
2
3 < x =) x > 9 =) x 62 A:
Thus, by Completeness, supA exists. Let
▯ = supA:
2
I claim that ▯ = 5; this will prove the claim.
2
Suppose ▯ < 5. Note that in this case, ▯ 2 A. (▯ is clearly positive, since it is an
upper bound of A and 2 2 A.) Since ▯ < 5, we have 5▯▯2 2 R. The Archimedean
Property therefore implies the existence of a natural number n such that
2▯ + 1
n > ;
5 ▯ ▯2
from which we deduce that
2▯ 1 2
+ < 5 ▯ ▯ :
n n
Since 2▯+ 1 ▯ 2▯+ 1 for all n 2 N, we see that
n n2 n n
2▯ 1
+ < 5 ▯ ▯ ;
n n2
or in other words, that
(▯ + 1=n) < 5:
But this means that ▯ + 1=n 2 A, contradicting that ▯ is an upper bound of A.
2 2▯
Next, suppose instead that ▯ > 5. Then ▯2▯5 2 R, so by Archimedean Proper

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