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Solutions to MATH105 exam September 2011
Section A
3 marks 1.a) For a real number x, x^2 + 2x − 3 = 0 if and only if x = 1 or x = −3. This is true because x^2 + 2x − 3 = (x − 1)(x + 3) = 0 ⇔ x − 1 = 0 or x + 3 = 0. 3 marks b) For a real number x, if x is greater than 2, then x is greater than 1 and less than 3. This is clearly false. For example if x = 4 then 4 > 2 but it is not true that 4 < 3. Standard home- work exercises 6 marks in total
1 mark 2a)∃x ∈ R, x 6 ∈ Q. 3 marks b) ∃x ∈ R, x < 3 ∧ x^2 ≥ 9. Standard home- work exercises 4 marks in total
1 mark 3a) No, 2 ∈/ [0, 2). 1 mark b) No 3 ∈/ X because 3^2 > 5. 1 mark c) No 1 ∈/ X. 1 mark d) Yes 1 mark e) No because 1 − 2 i is not a real number 1 mark f) No because
3 is not rational. Standard home- work exercises: no reasoning required. 6 marks in total
1 mark 4a) 1 < 3 x − 5 ⇔ 6 < 3 x ⇔ 2 < x.
2 marks b) If 1 + x > 0 then − 1 <
2 − x 1 + x
2 − x 1 + x
< 1 ⇔ − 1 − x <
2 − x < 1 + x ⇔ − 1 < 2 < 1 + 2x ⇔
< x, which is compatible with x > −1.
2 marks If 1 + x < 0 then− 1 <
2 − x 1 + x
< 1 ↔ − 1 − x > 2 − x >> 1 + x ⇒ − 1 > 1, which is never true. So altogether we have − 1 <↔
1 + x 1 − x
< x. It is permissible to do this question by sketching the graph. Standard home- work exercises. 5 marks in total
1 marks 5. To start the induction, 2^4 = 16 < 24 = 4!. So 2n^ < n! is true for n = 4.
5 marks
Now suppose inductively that n ≥ 4 and 2n^ < n!. Then
2 n+1^ = 2 · 2 n^ < 2 · n! < (n + 1) · n! = (n + 1)!
So true for n implies true for n + 1 and 2n^ < n! is true for all n ≥ 4. Standard home- work exercise 6 marks in total
- Performing integral row operations to implement Euclid’s algo- rithm:
R 2 − 2 R 1
R 1 − 2 R 2
R 2 − 3 R 1
4 marks
As a result of this: 1 mark (i) the g.c.d. d is 24; 1 mark (ii) from the last row of the last matrix, r = 7 and s = 17; 1 mark (iii) from the first row of either of the last two matrices m = 5 and n = −2; 2 marks (iv) The l.c.m is 408 × 7 = 2856. Standard home- work exercise 9 marks in total
3 marks 7a) f ((− 1 , ∞)) = (− 1 , ∞) because the cube root of x exists for all x ∈ R, f is increasing, f (−1) = −1 and f (x) → +∞ as x → +∞. So the image of f is (− 1 , , ∞) and f is surjective. Also, f is injective, because f is strictly increasing, and any strictly increasing function is injective.
4 marks b) f (x) = y ⇔ y =
x + 1 x − 1
⇔ xy − y = x + 1 ⇔ x(y − 1) = y + 1 ⇔
x =
y + 1 y − 1
. Now
y + 1 y − 1
is defined for y ∈ R ⇔ y 6 = 1. So the image of f is (−∞, 1) ∪ (1, ∞) 6 = R and f is not surjective. However, f is injective, because, for any y 6 = 1, the only value of x for which f (x) = y is x =
y + 1 y − 1
Standard home- work exercise 7 marks in total
Section B
Theory from lec- tures 4 marks
- ∼ is reflexive if x ∼ x∀x ∈ X ∼ is symmetric if
x ∼ y ⇒ y ∼ x ∀ x, y ∈ X.
∼ is Transitive if
(x ∼ y ∧ y ∼ z) ⇒ x ∼ z ∀ x, y ∈ X.
Standard home- work exercise 4 marks a) ∼ is reflexive because 3 | x − x = 0 for all x ∈ Z. It is symmetric because if x − y = 3m for some m ∈ Z then y − x = 3(−m). So ∼ is not an equivalence relation.It is also transitive because if x − y = 3m and y − z = 3p for m and p ∈ Z then x − z = 3(m + p) where m + p ∈ Z. Standard home- work exercise 2 marks
b) If x = 1 then x ∈ Z and xx = x^2 = 1 is not even. So ∼ is not symmetric and hence not an equivalence relation.
Standard home- work exercise 1 mark
c) If x =
then x ∈ Q and xx = x^2 =
∈/ Z. So once again ∼ is not symmetric and not an equivalence relation.
Unseen 4 marks
d) If z ∈ C then z − z = 0 = 0 + 0i. So ∼ is reflexive. If z − w = m + ni for m, n ∈ Z then w − z = −m + (−n)i and −m, −n ∈ Z So ∼ is symmetric. If z − w = m 1 + n 1 i and w − v = p + qi, where m, n, p, q ∈ Z, then z − v = (z − w) + (w − v) = m + ni + p + qi = (m + p) + (q + n)i and m + p, n + q ∈ Z. So ∼is transitive and ∼ is an equivalence relation. 15 marks in total
3 marks 11. We have x 0 = 2 and x 1 =
. So x^21 − 3 =
. So (i), (iii) and (iv) hold for n = 0 and (v) holds for n = 1.
2 marks (i) If xn > 0 then
x 2 2
and
2 xn
0 and xn+1 > 0
5 marks (ii)
x^2 n+1 − 3 =
xn 2
2 xn
x^2 n 4
4 x^2 n
x^2 n 4
4 x^2 n
4 x^2 n
(x^4 n − 6 x^2 n + 9)
(x^2 n − 3)^2 4 x^2 n
1 mark (iii) From (ii), we see that x^2 n+1 − 3 > 0
4 marks (iv) From (iii) for xnwe see that xn+1 − xn = −
x^2 n − 3 2 xn
< 0 and hence xn+1 < xn. From xn+1 > 0 (i) and x^2 n+1 − 3 > 0 (iii) we see that xn+1 > 1. Standard home- work exercises 15 marks in total
Theory from lec- tures 6 marks
13(i) A set A ⊂ Q is a Dedekind cut if
- A is nonempty, and bounded above,
- x ∈ A ∧ y < x ⇒ y ∈ A
- A does not have a maximal element.
Similar to home- work exercises medskip
1 mark (ii)a) {x ∈ Q : x > 1 } is not bounded above, so not a Dedekind cut 2 marks (ii)b)2 is a maximal element of {x ∈ Q : x ≤ 2 } and so A is not a Dedekind cut Similar to prac- tice exam
6 marks (iii)A is bounded above by 1, which is not in A, because x^2 +x−1 = (x + 12 )^2 − 54 is strictly increasing for x ≥ −^12 , and 1^2 + 1 − 1 > 0. If a ∈ A and b < a then if b < 0 we have b ∈ A. If 0 ≤ b < a then since x^2 + x − 1 is strictly increasing on [0, ∞), we have b^2 + b − 1 < a^2 + a − 1 < 0 and b ∈ A. If 0 < a, ε < 1 then
(a + ε)^2 + a + ε − 1 = a^2 + a − 1 + 2aε + ε^2 + ε < a^2 + a − 1 + 2ε + ε + ε
< a^2 + a − 1 + 4ε
If ε < −
a^2 + a − 1 4
then a^2 + a − 1+ 4ε < 0 and, if ε ∈ Q, a + ε ∈ A. So a is not maximal, for any a ∈ A, and A is a Dedekind cut. 15 marks in total
