Solutions to Problem set #3
Math 313
1. (a) If x ≤ 1, then since x is positive we have x2 ≤ 12 = 1 < 3, a contradiction since
x2 = 3. If x ≥ 1, then x2 ≥ 22 = 4 > 3, which is again a contradiction to x2 = 3.
Hence 1 < x < 2.
(b) Since 1 < x < 2, we have 1 < a/b < 2. Since b > 0 this implies b < a < 2b.
Adding b to both sides of each inequality, we get 0 < a − b < b.
(c) Since x2 = 3, we have (x − 1)(x + 1) = x2 − 1 = 2. Since x > 1 we have x − 1 6= 0;
hence we may divide both sides of the equality (x − 1)(x + 1) = 2 to obtain
2/(x − 1) = x + 1, which implies that 2/(x − 1) − 1 = x.
(d) We have
x=
2
2b
3b − a
2
−1=
−1=
−1=
.
x−1
(a/b) − 1
a−b
a−b
Since 3b − a and a − b are integers, and since a − b > 0 by (b), it now follows
that a − b is a denominator for x. But by (b) we have a − b < b, which is a
contradiction since b is the least denominator for x.
2. (a) If x ≥ 0 then |x| = x ≥ 0. If x < 0 then |x| = −x > 0. This proves the first
assertion. To prove the second, note that if x > 0 then −x < 0, so that |x| = x
and | − x| = −(−x) = x, and hence | − x| = |x| in this case. If x < 0 then −x > 0,
so that |x| = −x and | − x| = −x, and hence | − x| = |x| in this case as well. If
x = 0 then x = −x and hence |x| = | − x|.
(b) If x ≥ 0 then |x| = x, and if x < 0 then |x| = −x. Thus in either case we have
|x| = ±x.
(c) If x ≥ 0 then |x| = x, so that in particular x ≤ |x|. If x < 0 then |x| = −x > 0.
Hence in this case we have x < |x|, so that in particular x ≤ |x|.
(d) By (c) we have x ≤ |x| and y ≤ |y|, and hence x + y ≤ |x| + |y|. Substituting
−x for x and −y for y in this last inequality, we obtain −(x + y) = (−x) +
(−y) ≤ | − x| + | − y|; since | − x| = |x| and | − y| = |y| by (a), it follows that
−(x + y) ≤ |x| + |y|. Thus each of the numbers x + y and −(x + y) is less than or
equal to |x| + |y|. But by (b), |x + y| is equal to either x + y or −(x + y). Hence
in either case we have |x + y| ≤ |x| + |y|.
(e) By (b) we may write |x| = ux, where u is either 1 or −1. Likewise, we may write
we may write |x| = vx, where v is either 1 or −1. Hence |x||y| = uvxy. Since
each of the numbers u and v is either 1 or −1, their product is also either 1 or −1;
hence |x||y| = ±(xy). By (b) we also have |xy| = ±(xy). Hence |xy| = ±(|x||y|).
If |xy| = +(|x||y|), the assertion is proved. Now suppose that |xy| = −(|x||y|).
According to (a), we have |x| ≥ 0 and |y| ≥ 0; hence |x||y| ≥ 0 and therefore
−(|x||y|) ≤ 0. But (a) also gives |xy| ≥ 0, so that the equality |xy| = −(|x||y|)
1
implies −(|x||y|) ≥ 0. Hence −(|x||y|) = 0, which now implies |x||y| = 0 and
|xy| = 0, so that |x||y| = |xy| in this case as well.
3. (a) The base case is y1 > r, which is included in the hypothesis 0 < y1 − r < 1/2.
For the induction step, assume that yn > r for a given n. Since xn = a/yn
and yn+1 = (xn + yn )/2, we may apply the theorem with y = yn , x = xn and
y 0 = yn+1 . The first inequality of the theorem then becomes 0 < yn+1 − r, i.e.
yn+1 > r. This completes the induction.
(b) Since (a) gives yn > r > 0, we have xn = a/yn < a/r = r2 /r = r.
(c) We have yn > r by (a) and xn < r by (b). Hence xn < yn . Therefore yn+1 =
(xn + yn )/2 < (yn + yn )/2 = yn .
(d) By induction on a natural number n we wish to show that yn < a. The base case
is y1 < a, which is part of the hypothesis. For the induction step, suppose that
yn < a for a given n. According to (c) we have yn+1 < yn , so that yn+1 < a by
transitivity. This completes the induction.
For any n, since yn < a, we have xn = a/yn > a/a = 1.
Since yn > r, and since xn = a/yn and yn+1 = (xn + yn )/2, we may apply the
theorem with y = yn , x = xn and y 0 = yn+1 . This gives
1
(yn − r)2 .
0 < yn−1 − r <
2xn
Since xn > 1 we have (yn − r)2 /(2xn ) < (yn − r)2 /2, and hence
1
0 < yn+1 − r < (yn − r)2 .
2
(e) We use induction on n. The base case is
1
,
2
which is part of the hypothesis. For the induction step, suppose that for a given
n we have
1
0 < yn − r < 2n −1 .
2
Then by the last assertion of (d), we have
2
1
1
0 < yn+1 − r <
2 22n −1
1
1
=
2 22(2n −1)
1
= 1+2(2n −1)
2
1
= 1+(2n+1 −2)
2
1
= 2n+1 −1 ,
2
0 < y1 − r <
2
which completes the induction.
(f ) If r ≤ 5/2 then
r2 ≤ (5/2)2 = 6
1
< 7,
4
a contradiction since r2 = 7. Likewise, if r ≥ 3 then r2 ≥ 32 = 9 > 7, a
contradiction. Hence 5/2 < r < 3. If we set y1 = 3 it follows that y1 − 3 <
y1 − r < y1 = 5/2, i.e. that 0 < y1 − r < 1/2. We also have y1 < a since a = 7
an y1 = 3. Thus the hypotheses stated in (a) hold. By the result of (e), taking
n = 4, we have
1
1
1
0 < y4 − r < 24 −1 = 15 <
.
2
30, 000
2
Thus the required inequality holds if we set Y = y4 . Using the recursive formulae
xn = 7/yn and yn+1 = (xn + yn )/2, and recalling that y1 = 3, we find that
x1 = 7/3,
1 7
8
7
21
y2 =
+3 = ,
x2 =
=
,
2 3
3
8/3
8
1 21 8
127
7
336
y3 =
+
=
,
x3 =
=
,
2 8
3
48
127/48
127
1 336 127
32257
y4 =
+
=
.
2 127
48
12192
Thus we can take Y = 32257/12192.
4. First suppose that M ∈ X. Since M is an upper bound for X, we have M ≥ x for
every x ∈ X, so that M is a greatest element of X. Conversely, suppose that X has
a greatest element, say x0 . Then x0 ≥ x for every x ∈ X, so that x0 is an upper
bound for X. If L is any upper bound for X, we have L ≥ x for every x ∈ X, and in
particular L ≥ x0 . Thus x0 is a least upper bound for X. By the uniqueness of least
upper bounds, we have x0 = M . But x0 is in particular an element of X, and hence
M ∈ X.
3