Download HOMEWORK ASSIGNMENT 2 Exercise 1 ([1, Exercise 3.4]). Without

HOMEWORK ASSIGNMENT 2
ACCELERATED PROOFS AND PROBLEM SOLVING [MATH08071]
Each problem will be marked out of 4 points.
Exercise 1 ([1, Exercise 3.4]). Without using a calculator, find the cube root of 2, correct
to 1 decimal place.
√
Solution. The decimal representation of 3 2 looks like
√
3
2 = n.a1 a2 a3 . . . ,
where n is a positive integer, and each ai is a digit in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}). Note that
this representation means an infinite sum
a1
a2
a3
n+
+ 2 + 3 + ··· ,
10 10
10
√
which means, in particular, that n is the biggest non-negative integer such that n 6 3 2.
Since 1 < 2 < 8, we have
√
√
√
3
3
3
1 = 1 < 2 < 8 = 2,
√
which implies that 0 < 3 2 − 1 < 1. Hence, we have n = 1. Then
√
a2
a3
3
0 < 10 2 − 1 = a1 .a2 a3 . . . = a1 +
+ 2 + ··· ,
10 10
√
√
so a1 is is the biggest non-negative integer such that a1 6 10( 3 2 − 1) = 3 2000 − 10.
Note that 123 = 1728 and 133 = 2197. Thus, we have
√
√
√
3
3
3
12 = 1728 < 2000 < 2197 = 13,
√
3
which implies
that
2
<
2000 − 10 < 3. Thus, we have a1 = 2, which is enough for our
√
3
purposes: 2 ≈ 1.2. But we can easily proceed further to find a2 . We have
!
√
√
a3
a4
3
3
0 < 2000000 − 120 = 10 10 2 − 1 − 2 = a2 .a3 . . . = a2 +
+ 2 + ··· ,
10 10
√
and a2 is the biggest non-negative integer such that a2 6 3 2000000 − 120. Since
we see that 5 <
√
3
1252 = 1953125 < 20000 < 2000376 = 1262 ,
2000000−120 < 6, which implies that a2 = 5. Thus, we have
√
3
2 ≈ 1.25.
Exercise 2 ([1, Exercise 4.2]). For which values of x is
x2 + x + 1 >
x−1
.
2x − 1
Solution. Let us solve this inequality in a simple (but may be boring) way. Namely, we
consider two cases: 2x − 1 > 0 and 2x − 1 < 0.
Suppose that 2x − 1 > 0, which is the same as assuming x > 1/2. Then
x−1
x2 + x + 1 >
⇐⇒ (2x − 1) x2 + x + 1 > x − 1 ⇐⇒ 2x3 + x2 > 0,
2x − 1
This assignment is due on Thursday 8th October 2015.
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where 2x3 + x2 = x2 (2x + 1). Since x 6= 0, we see that
x−1
1
x2 + x + 1 >
⇐⇒ x2 (2x + 1) > 0 ⇐⇒ 2x + 1 > 0 ⇐⇒ x > − ,
2x − 1
2
but we should keep in mind that x > 1/2 by assumption. Thus, we see that the required
inequality holds for every x > 1/2.
Suppose that 2x − 1 < 0, which is the same as assuming x < 1/2. Then
x−1
x2 + x + 1 >
⇐⇒ (2x − 1) x2 + x + 1 6 x − 1 ⇐⇒ 2x3 + x2 6 0,
2x − 1
3
2
2
where 2x + x = x (2x + 1). If x = 0, then
x−1
= 1,
1 = x2 + x + 1 >
2x − 1
which implies that x = 0 is a solution. If x 6= 0, then
1
x−1
x2 + x + 1 >
⇐⇒ x2 (2x + 1) 6 0 ⇐⇒ 2x + 1 > 0 ⇐⇒ x > − ,
2x − 1
2
which implies that the required inequality must hold for every x 6 −1/2 and does not
hold if −1/2 < x < 1/2 and x 6= 0.
Thus, we see that the inequality
x−1
x2 + x + 1 >
2x − 1
holds if and only if either x 6 −1/2, or x = 0, or x > 1/2. This can be seen from the
picture
,
where the red curve is the graph of the function x2 + x + 1, and the blue curve is the
graph of the function (x − 1)/(2x − 1). Ignore the vertical blue line x = 1/2 (the function
(x − 1)/(2x − 1) is undefined at this point).
Exercise 3 ([1, Exercise 4.4]). (a) Find the set of real numbers x 6= 0 such that 2x + x1 <
3.
2
(b) Find the set of real numbers t such that the equation x2 + tx + 3 = 0 has two distinct
real solutions.
Solution. Lets first find the set of real numbers x 6= 0 such that 2x + 1/x < 3. If x < 0,
then 2x + 1/x < 0 < 3. If x > 0, then
2x +
1
< 3 ⇐⇒ 2x2 + 1 < 3x ⇐⇒ 2x2 − 3x + 1 < 0 ⇐⇒ (x − 1)(x − 1/2) < 0,
x
because x > 0. But (x − 1)(x − 1/2) < 0 if and only if 1/2 < x < 1. Thus, we see that
2x + 1/x < 3 if and only if either x < 0 or 1/2 < x < 1. This can be seen from the picture
,
where the red curve is the graph of the function 2x + 1/x − 3.
Now let us find the set of real numbers t such that the equation x2 + tx + 3 = 0 has two
distinct real solutions. Note that
x2 + tx + 3 = 0 ⇐⇒
x+
t2
t 2
+3−
= 0 ⇐⇒
2
4
x+
t 2 t2 − 12
=
,
2
4
which implies that the equation x2 + tx + 3 = 0 has two distinct real solutions if and only
if t2 > 12. Indeed, if t2 < 12 (t2 = 12, respectively), then the equation x2 + tx + 3 = 0 has
no real solutions (one real solution, respectively). This can be seen from the picture
3
,
2 + 3x + 3, the blue curve is the graph of
where the red curve
is
the
graph
of
the
function
x
√
the function x2 + 12x + 3, and the green curve is the graph of the function x2 + 4x + 3.
2
Thus,√the equation
√ x + tx + 3 = 0 has two distinct real solutions if and only if either
t < − 12 or t > 12.
Exercise 4 ([1, Exercise 5.4]). Find an integer n and a rational t such that
1
1
nt = 2 2 3 3 .
Solution. Let n be a positive integer and t be a rational. Then
!6
6
1 1
1
1 1
⇐⇒ n6t = 23 32 ⇐⇒ nt = 72 6 ,
nt = 2 2 3 3 ⇐⇒ nt = 2 2 3 3
which implies that we can put n = 72 and 1/6. Note that the answer is not unique, since
we can also put n = 722 = 5184 and t = 1/12 etc.
Exercise 5 ([1, Exercise 5.6]). Find all real solutions of the equation
1
1
x 2 − (2 − 2x) 2 = 1.
Solution. Note that the equation can be rewritten as
√
√
x − 2 − 2x = 1,
and the
√ of such real x implies that x > 0 and x 6 1, since we implicitly assume
√ existence
that x and 2 − 2x are real as well. Then
√
√
√
x − 2 − 2x < 1 = 1
√
√
if x < 1, √
and √
x − 2 − 2x = 1 if x = 1. Thus, x = 1 is the only real solutions of the
equation x − 2 − 2x = 1.
References
[1] M. Liebeck, A concise introduction to pure mathematics
Third edition (2010), CRC Press
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