Download Real Analysis 1, Exercise Sheet 4, October 30 2012

Real Analysis 1, Exercise Sheet 4, October 30 2012
Question 1. Using the axioms of an ordered field, prove that:
(i) If z 6= 0 and xz = yz, then x = y.
(ii) The additive neutral element, denoted 0, is unique.
(iii) For each x 6= 0, there is a unique x1 .
(iv) If 0 < x < y, then x2 < y 2 .
[The axioms are listed on the reverse of the sheet to help you.]
Question 2. Which of the following sets satisfy the axioms of an ordered
field, and if not, which axiom(s) fail(s)?
Z
[0, ∞)
{transcendental numbers}
Z/5
where Z/5 = {0, 1, 2, 3, 4} with addition and multiplication mod 5 (so
3 + 4 ≡ 2 and 3 · 4 ≡ 2).
Question 3. Put the correct inequality sign (i.e., < or >) in place of the
•:
1
• n12 .
a) Assuming n + 10 < n2 : n+10
b) Assuming n + 10 < n2q
: ln(n + 10)
q • 2 ln n.
1
•
y+1 .
√
√
d) Assuming 0 < x < y < 1: tan x • tan y.
2
2
e) Assuming 0 < x < y: sin(e−1/x ) • sin(e−1/y ).
c) Assuming 0 < x < y:
1
x+1
1
The Axioms of an Ordered Field
A field is closed
Addition
∀x, y ∈ R x + y ∈ R
Multiplication
∀x, y ∈ R xy ∈ R
Associative law
∀x, y, z ∈ R (x + y) + z = x + (y + z)
∀x, y, z ∈ R (xy)z = x(yz)
Commutative law
∀x, y ∈ R x + y = y + x
∀x, y, z ∈ R xy = yx
Neutral element
∃ n1 ∈ R ∀x ∈ R x + n1 = x
∃ n2 ∈ R n2 6= n1 ∀x ∈ R xn2 = x
Negative and
reciprocal
∀x ∈ R ∃(−x) ∈ R x + (−x) = n1
∀x ∈ R x 6= n1 ∃ x1 ∈ R x
Distributive law
∀x, y, z ∈ R x(y + z) = xy + xz
Order is complete
∀x, y ∈ R either x < y or x > y or x = y
Transitivity
∀x, y, z ∈ R x < y and y < z implies x < z
∀x, y, z ∈ R
x < y implies x + z < y + z
2
1
x
∀x, y, z ∈ R with z > 0
x < y implies xz < yz
= n2
Solutions to RA1 Exx Sheet 4, 2012:
Question 1.
(i) If z 6= 0 and xz = yz then
1
z
1
x(z )
z
x.1
x
(xz)
1
z
1
= y(z )
z
= y.1
= y.
= (yz)
The axioms needed are clear from the changes made at each step.
(ii) The additive neutral element, denoted 0, is unique:
Suppose 0̃ is another additive neutral element. Then
0 = 0 + 0̃
because 0̃ is additive neutral, and
0̃ = 0̃ + 0
because 0 is additive neutral. Since addition is commutative, 0 + 0̃ = 0̃ + 0,
whence 0 = 0̃.
(iii) For each x 6= 0, there is a unique x1 .
Suppose t is another possible x1 . Then
tx = 1
1
1
(tx) = 1.
x
x
1
1
t(x ) =
x
x
1
t.1 =
x
1
t = .
x
Again, axioms used should be clear from line changes.
(iv) If 0 < x < y, then x2 < y 2 .
Note that x < y and x > 0 together imply (using the axioms) x2 < xy,
3
while x < y and y > 0 together imply (similarly) xy < y 2 . By transitivity,
it follows that x2 < y 2 .
Question 2. Which of the following sets satisfy the axioms of an ordered
field, and if not, which axiom(s) fail(s)?
Z
[0, ∞)
{transcendental numbers}
Z/5
where Z/5 = {0, 1, 2, 3, 4} with addition and multiplication mod 5 (so 3 +
4 ≡ 2 and 3 · 4 ≡ 2).
does not satisfy the axioms relating to multiplicative inverses: e.g. 2 ∈ Z,
but no element z ∈ Z satisfies 2z = 1.
Z
[0, ∞) fails to contain additive inverses: consider 1, for example: no element
t of [0, ∞) is such that t + 1 = 0.
The transcendental numbers do not contain a 0 or a 1: both these numbers
are algebraic.
Z/5
=: Z5 does not have an order relation. To see this, suppose for a
contradiction that ≺ is an order relation on Z5 . Then, since 2 6= 3, either
2 ≺ 3 or 3 ≺ 2.
Using only addition: Suppose 2 ≺ 3. Then 2 + 1 ≺ 3 + 1 gives 3 ≺ 4;
3 + 1 ≺ 4 + 1 gives 4 ≺ 0. Continuing in this way, 1 ≺ 2 ≺ 3 ≺ 4 ≺ 0 ≺ 1,
which is a contradiction (we suppose x ≺ x is impossible). If 3 ≺ 2 then
all ≺ are reversed, giving the same contradiction.
Using only multiplication: Suppose to begin with that 2 ≺ 3.
(i) If 2 0 and 3 0 then 2.2 ≺ 2.3 and 2.3 ≺ 3.3. In Z5 , this reads
4 ≺ 1 and 1 ≺ 4. But at most one of these can be true (by trichotomy),
which is a contradiction.
(ii) If 2 ≺ 0 and 3 ≺ 0 then 2.2 2.3 and 2.3 3.3, which gives the same
contradiction as case (i).
(iii) If 2 0 and 3 ≺ 0 then 2 3, contradicting the asssumption 2 ≺ 3.
(iv) If 2 ≺ 0 and 3 0 then 4 1 results from both 2.2 2.3 and
2.3 ≺ 3.3. But then 4.3 > 1.3, implying 2 3, and giving the same
contradiction as in (iii).
4
If 3 ≺ 2 then a similar argument leads to a contradiction. Either way,
there can be no order relation on Z5 . Note that Z5 does satisfy the field
axioms (i.e., those not involving the order relation).
Question 3.
1
a) Assuming n + 10 < n2 : n+10
> n12 .
b) Assuming n + 10 < n2q
: ln(n + q
10) < 2 ln n.
c) Assuming 0 < x < y:
1
x+1
>
1
y+1 .
√
√
d) Assuming 0 < x < y < 1: tan x < tan y.
2
2
e) Assuming 0 < x < y: sin(e−1/x ) < sin(e−1/y ).
5