MATH 2200 – Problem Set #1 Solutions
3 Feb. 2017
Question 3: prove the following...
a)
−(−x) = x
Proof.
−x + −(−x) = 0
− x + −(−x) + x = 0 + x
[A.5]
(−x + x) + −(−x) = x + 0
[A.2, A.3]
[A.1]
0 + −(−x) = x
[A.5]
−(−x) = x
[A.4]
Alternatively:
Proof. (Using uniqueness of the additive inverse.) ∀x ∈ R we have both
(−x) + x = 0
[A.5]
and
(−x) + −(−x) = 0
[A.5 applied to y = −x.]
So both −(−x) and x are additive inverses for −x. Thus, by uniqueness of the additive inverse, we
must have
x = −(−x).
b)
(i)
(−x) · y = −(x · y)
Proof.
−x + x = 0
[A.5]
(−x + x) · y = 0 · y
[M.1]
(−x) · y + x · y = 0
[D.L., Thm. ( x · 0 = 0)]
((−x) · y + x · y) + − (x · y) = 0 + − (x · y)
(−x) · y + (x · y + −(x · y)) = − (x · y)
[A.1]
[A.2, A.3, A.4]
(−x) · y + 0 = − (x · y)
[A.5]
(−x) · y = − (x · y)
[A.4]
1
(−x) · (−y) = x · y
(ii)
Proof.
(−x) · (−y) = − (x · (−y))
by (i) above with y → −y
(−x) · (−y) = − ((−y) · x)
M.2
(−x) · (−y) = (−(−y)) · x
by (i) again, in reverse
(−x) · (−y) = y · x
[3a]
(−x) · (−y) = x · y
M.2
c)
(i)
If x 6= 0 then
1
x
6= 0.
Proof. By contradiction. . . Suppose to the contrary that x 6= 0 and
1
·x=1
x
0·x=1
0=1
1
x
= 0. Then
[M.5]
[by assumption x = 0]
[Thm. (0 · x = 0)]
This contradicts [M.4]. Thus
x 6= 0 =⇒
1
6= 0
x
(ii)
If x 6= 0 then 1/ x1 = x.
Proof.
1
1
1/
· =1
x
x
1 1
·x=1·x
1/ ·
x x
1
1
1/
·
·x =x
x
x
1
1/
·1=x
x
1
1/ = x
x
[M.5]
[M.1]
[M.3, M.4]
[M.5]
[M.4]
2
Alternatively:
Proof. (Using uniqueness of the multiplicative inverse.) ∀x ∈ R, x 6= 0, we have both
1
·x=1
x
and
1
1
·
=1
x 1/x
[M.5]
[M.5 applied to y = 1/x].
1
are multiplicative inverses for x1 (i.e. numbers which multiplied by
So both x and 1/x
Thus, by uniqueness of the multiplicative inverse, we must have
x=
1
x
give 1).
1
.
1/x
d)
If x · z = y · z and z 6= 0 then x = y.
Proof. Suppose x · z = y · z and z 6= 0. Then
1
z
exists and
1
1
(x · z) · = (y · z) ·
z
z
1
1
=y· z·
x· z·
z
z
x·1=y·1
[M.5]
x=y
[M.4]
[M.1, M.5]
[M.3]
e)
If x 6= 0 then x2 > 0.
Proof. Suppose x 6= 0. Then either x > 0 or x < 0 [O.1]. There are two cases to consider:
case x > 0:
x>0
x·x>0·x
x2 > 0
[by assumption]
[O.4]
[Thm. (0 · x = 0)]
case x < 0:
x<0
x·x>0·x
x2 > 0
[by assumption]
[Thm. 3.2.2g]
[Thm. (0 · x = 0)]
3
f)
0<1
Proof.
0 6= 1
0<1·1
0<1
[M.4]
[Thm. 3(e) above]
[M.4]
Alternatively:
Proof. (By contradiction.) Suppose on the contrary that 0 ≥ 1. Since 0 6= 1 [M4] this implies 0 > 1
[O1]. Suppose x ∈ R and x > 0. Then
1
·x=1<0
x
1
x·
·x <x·0
x
1
x·
·x<0
x
1·x<0
[M.5]
x<0
[M.4]
[M.5]
[M.1]
[M.3, Thm.3.2.2(b)]
which contradicts O1. Therefore 0 < 1.
g)
If x > 1 then x2 > x.
Proof. Suppose x > 1. Then x > 0 by 3(f) above [O.2]. Thus
x>1
[by assumption]
x·x>1·x
[O.4]
x2 > x
[M.4]
h)
If 0 < x < 1 then x2 < 1.
4
Proof. Suppose 0 < x < 1. Then
0<x<1
[by assumption]
0·x<x·x<1·x
[O.4]
2
[M.4] and Thm. (0 · x = 0)
0<x <x
2
x <x<1
[by assumption]
x2 < 1
[O.2]
i)
(I)
If x > 0 then
1
x
> 0.
Proof. By contradiction. Suppose to the contrary that x > 0 and
x·
1
<0
x
[O.1]
1
<x·0
x
[O.4]
1<0
1
x
≤ 0. Then
1
x
[M.5, Thm. (x · 0 = 0)]
This contradicts Thm. 3(f).
(II)
If x < 0 then
1
x
< 0.
Proof. By contradiction. Suppose to the contrary that x < 0 and
1
>0
x
x·
1
<x·0
x
1<0
1
x
≥ 0. Then
If 0 < x < y then 0 <
1
y
1
x
6= 0 [3(c)] so that
[O.1]
[3.2.2(g)]
[M.5, Thm. (x · 0 = 0)]
This contradicts Thm. 3(f).
j)
6= 0 [3(c)] so that
< x1 .
Proof.
0<x<y
1 1
1 1
1 1
0· · <x· · <y· ·
x y
x y
x y
1
1
0<1· <1·
y
x
0<
1
1
<
y
x
5
[by assumption]
[3i, O.2, O.4]
[Thm., M.5, O.2]
[M.4, O.2]
k)
If x · y > 0 then either (i) x > 0 ∧ y > 0, or (ii) x < 0 ∧ y < 0.
Proof. Suppose x · y > 0. Then both x 6= 0 and y 6= 0 (since otherwise x · y = 0 by prev. Thm.
(∀x, x · 0 = 0) which contradicts x · y > 0). There are two cases to consider:
case x > 0:
1
>0
x
1
· (x · y) > 0 · (x · y)
x
1·y >0
[3(i)]
[O.4, assumption x · y > 0]
[M.2, M.3, Thm.]
y>0
[M.5]
1
<0
x
[3(i)]
Hence x > 0 ∧ y > 0.
case x < 0:
1
· (x · y) < 0 · (x · y)
x
1·y <0
y<0
[O.4, assumption x · y > 0]
[M.2, M.3, Thm.]
[M.5]
Hence x < 0 ∧ y < 0.
Thus, either x > 0 and y > 0, or else x < 0 and y < 0.
l)
If 0 < x < y then xn < y n , ∀n ∈ N.
Proof. By induction. Suppose 0 < x < y. The base case (n = 1) is trivial:
0<x<y
1
x <y
[by assumption]
1
Next, the inductive argument: suppose that xn < y n for some particular n ∈ N. We show that
xn+1 < y n+1 :
xn < y n
n
[by assumption]
n
n
[O.4, assumption x < y]
n
n
[O.2]
x ·x<y ·x<y ·y
x ·x<y ·y
x
n+1
<y
n+1
Thus xn < y n , ∀n ∈ N.
6
√
√
If 0 < x < y then 0 < x < y.
√
Note that . is non-negative by definition.
m)
Proof. Suppose 0 < x < y. Then
x−x<y−x
[A.1, O.2]
y − x > 0.
Also
√
x 6= 0 and
√
y 6= 0 (since otherwise x = 02 = 0 contradicts x > 0), hence
√
√
x > 0, y > 0
[O.1]
√
√
√
x+ y >0+ y >0+0=0
[O.3, A.4]
√
√
x+ y >0
[O.2]
1
√
[3(i)]
√ >0
x+ y
Since both
y−x>0
and
√
1
√ >0
y+ x
we have
√
Thus
√
y>
y−
√
1
√ >0
x = (y − x) · √
y+ x
√
√
√
√
y− x+ x>0+ x
√
√
y> x
√
x > 0.
[O.4]
[O.3]
[A.4]
Question 4
If x ≥ 0 and ∀ε > 0, x ≤ ε then x = 0.
Proof. By contradiction. . . suppose to the contrary that:
• x≥0
• ∀ε > 0, x ≤ ε
• x 6= 0.
Thus x > 0 [O.1]. We claim ∃ε > 0 with x > ε, leading to a contradiciton. Indeed, let ε = x/2.
Then we have
x
0+x
x+x
<
= x.
0<ε= =
2
2
2
Thus x > ε [O.2], which contradicts the assumption that x ≤ ε.
7