A concept we will encounter over and over is the concept of absolute value. You want to think
of the absolute value as the “size” of a real number. Let us give a formal definition.
jxj :=
(
x if x 0;
x if x < 0:
Let us give the main features of the absolute value as a proposition.
Proposition 1.3.1.
(i) jxj 0, and jxj = 0 if and only if x = 0.
(ii) jxj = jxj for all x 2 R.
(iii) jxyj = jxj jyj for all x;y 2 R.
(iv) jxj2 = x2 for all x 2 R.
(v) jxj y if and only if y x y.
(vi) jxj x jxj for all x 2 R.
Proof. (i): This statement is obvious from the definition.
(ii): Suppose that x > 0, then jxj = (x) = x = jxj. Similarly when x < 0, or x = 0.
(iii): If x or y is zero, then the result is obvious. When x and y are both positive, then jxj jyj = xy.
xy is also positive and hence xy = jxyj. Finally without loss of generality assume that x > 0 and
y < 0. Then jxj jyj = x(y) = (xy). Now xy is negative and hence jxyj = (xy).
(iv): Obvious if x = 0 and if x > 0. If x < 0, then jxj2 = (x)2 = x2.
(v): Suppose that jxj y. If x > 0, then x y. Obviously y 0 and hence y 0 < x so
y x y holds. If x < 0, then jxj y means x y. Negating both sides we get x y. Again
y 0 and so y 0 > x. Hence, y x y. If x = 0, then as y 0 it is obviously true that
y 0 = x = 0 y.
On the other hand, suppose that y x y is true. If x 0, then x y is equivalent to jxj y.
If x < 0, then y x implies (x) y, which is equivalent to jxj y.
(vi): Just apply (v) with y = jxj.
A property used frequently enough to give it a name is the so called triangle inequality.
Proposition 1.3.2 (Triangle Inequality). jx+yj jxj+jyj for all x;y 2 R.
32 CHAPTER 1. REAL NUMBERS
Proof. From Proposition 1.3.1 we have jxj x jxj and jyj y jyj. We add these two
inequalities to obtain
(jxj+jyj) x+y jxj+jyj :
Again by Proposition 1.3.1 we have that jx+yj jxj+jyj.
There are other versions of the triangle inequality that are applied often.
Corollary 1.3.3. Let x;y 2 R
(i) (reverse triangle inequality)
jxjjyj
jxyj.
(ii) jxyj jxj+jyj.
Proof. Let us plug in x = ab and y = b into the standard triangle inequality to obtain
jaj = jab+bj jabj+jbj :
or jajjbj jabj. Switching the roles of a and b we obtain or jbjjaj jbaj = jabj. Now
applying Proposition 1.3.1 again we obtain the reverse triangle inequality.
The second version of the triangle inequality is obtained from the standard one by just replacing
y with y and noting again that jyj = jyj.
Corollary 1.3.4. Let x1;x2; : : : ;xn 2 R. Then
jx1+x2+ +xnj jx1j+jx2j+ +jxnj :
Proof. We will proceed by induction. Note that it is true for n = 1 trivially and n = 2 is the standard
triangle inequality. Now suppose that the corollary holds for n. Take n+1 numbers x1;x2; : : : ;xn+1
and compute, first using the standard triangle inequality, and then the induction hypothesis
jx1+x2+ +xn+xn+1j jx1+x2+ +xnj+jxn+1j
jx1j+jx2j+ +jxnj+jxn+1j:
Let us see an example of the use of the triangle inequality.
Example 1.3.5: Find a number M such that jx29x+1j M for all 1 x 5.
Using the triangle inequality, write
jx29x+1j jx2j+j9xj+j1j = jxj2+9jxj+1:
It is obvious that jxj2+9jxj+1 is largest when jxj is largest. In the interval provided, jxj is largest
when x = 5 and so jxj = 5. One possibility for M is
M = 52+9(5)+1 = 71:
There are, of course, other M that work. 71 is in fact much higher than it need be. But we didn’t ask
for the best possible M, just one that works.
1.3. ABSOLUTE VALUE 33
The last example leads us to the concept of bounded functions.
Definition 1.3.6. Suppose f : D!R is a function. We say f is bounded if there exists a number
M such that j f (x)j M for all x 2 D.
In the example we have shown that x29x+1 is bounded when considered as a function on
D = fx j 1 x 5g. On the other hand, if we consider the same polynomial as a function on the
whole real line R, then it is not bounded.
If a function f : D!R is bounded, then we can talk about its supremum and its infimum. We
write
sup
x2D
f (x) := sup f (D);
inf
x2D
f (x) := inf f (D):
To illustrate some common issues, let us prove the following proposition.
Proposition 1.3.7. If f : D!R and g: D!R are bounded functions and
f (x) g(x) for all x 2 D;
then
sup
x2D
f (x) sup
x2D
g(x) and inf
x2D
f (x) inf
x2D
g(x): (1.1)
You should be careful with the variables. The x on the left side of the inequality in (1.1) is
different from the x on the right. You should really think of the first inequality as
sup
x2D
f (x) sup
y2D
g(y):
Let us prove this inequality. If b is an upper bound for g(D), then f (x) g(x) b and hence b is
an upper bound for f (D). Therefore taking the least upper bound we get that for all x
f (x) sup
y2D
g(y):
But that means that supy2D g(y) is an upper bound for f (D), hence is greater than or equal to the
least upper bound of f (D).
sup
x2D
f (x) sup
y2D
g(y):
The second inequality (the statement about the inf) is left as an exercise.
Do note that a common mistake is to conclude that
sup
x2D
f (x) inf
y2D
g(y): (1.2)
The inequality (1.2) is not true given the hypothesis of the claim above. For this stronger inequality
you need the stronger hypothesis
f (x) g(y) for all x 2 D and y 2 D.
The proof is left as an exercise.
of the absolute value as the “size” of a real number. Let us give a formal definition.
jxj :=
(
x if x 0;
x if x < 0:
Let us give the main features of the absolute value as a proposition.
Proposition 1.3.1.
(i) jxj 0, and jxj = 0 if and only if x = 0.
(ii) jxj = jxj for all x 2 R.
(iii) jxyj = jxj jyj for all x;y 2 R.
(iv) jxj2 = x2 for all x 2 R.
(v) jxj y if and only if y x y.
(vi) jxj x jxj for all x 2 R.
Proof. (i): This statement is obvious from the definition.
(ii): Suppose that x > 0, then jxj = (x) = x = jxj. Similarly when x < 0, or x = 0.
(iii): If x or y is zero, then the result is obvious. When x and y are both positive, then jxj jyj = xy.
xy is also positive and hence xy = jxyj. Finally without loss of generality assume that x > 0 and
y < 0. Then jxj jyj = x(y) = (xy). Now xy is negative and hence jxyj = (xy).
(iv): Obvious if x = 0 and if x > 0. If x < 0, then jxj2 = (x)2 = x2.
(v): Suppose that jxj y. If x > 0, then x y. Obviously y 0 and hence y 0 < x so
y x y holds. If x < 0, then jxj y means x y. Negating both sides we get x y. Again
y 0 and so y 0 > x. Hence, y x y. If x = 0, then as y 0 it is obviously true that
y 0 = x = 0 y.
On the other hand, suppose that y x y is true. If x 0, then x y is equivalent to jxj y.
If x < 0, then y x implies (x) y, which is equivalent to jxj y.
(vi): Just apply (v) with y = jxj.
A property used frequently enough to give it a name is the so called triangle inequality.
Proposition 1.3.2 (Triangle Inequality). jx+yj jxj+jyj for all x;y 2 R.
32 CHAPTER 1. REAL NUMBERS
Proof. From Proposition 1.3.1 we have jxj x jxj and jyj y jyj. We add these two
inequalities to obtain
(jxj+jyj) x+y jxj+jyj :
Again by Proposition 1.3.1 we have that jx+yj jxj+jyj.
There are other versions of the triangle inequality that are applied often.
Corollary 1.3.3. Let x;y 2 R
(i) (reverse triangle inequality)
jxjjyj
jxyj.
(ii) jxyj jxj+jyj.
Proof. Let us plug in x = ab and y = b into the standard triangle inequality to obtain
jaj = jab+bj jabj+jbj :
or jajjbj jabj. Switching the roles of a and b we obtain or jbjjaj jbaj = jabj. Now
applying Proposition 1.3.1 again we obtain the reverse triangle inequality.
The second version of the triangle inequality is obtained from the standard one by just replacing
y with y and noting again that jyj = jyj.
Corollary 1.3.4. Let x1;x2; : : : ;xn 2 R. Then
jx1+x2+ +xnj jx1j+jx2j+ +jxnj :
Proof. We will proceed by induction. Note that it is true for n = 1 trivially and n = 2 is the standard
triangle inequality. Now suppose that the corollary holds for n. Take n+1 numbers x1;x2; : : : ;xn+1
and compute, first using the standard triangle inequality, and then the induction hypothesis
jx1+x2+ +xn+xn+1j jx1+x2+ +xnj+jxn+1j
jx1j+jx2j+ +jxnj+jxn+1j:
Let us see an example of the use of the triangle inequality.
Example 1.3.5: Find a number M such that jx29x+1j M for all 1 x 5.
Using the triangle inequality, write
jx29x+1j jx2j+j9xj+j1j = jxj2+9jxj+1:
It is obvious that jxj2+9jxj+1 is largest when jxj is largest. In the interval provided, jxj is largest
when x = 5 and so jxj = 5. One possibility for M is
M = 52+9(5)+1 = 71:
There are, of course, other M that work. 71 is in fact much higher than it need be. But we didn’t ask
for the best possible M, just one that works.
1.3. ABSOLUTE VALUE 33
The last example leads us to the concept of bounded functions.
Definition 1.3.6. Suppose f : D!R is a function. We say f is bounded if there exists a number
M such that j f (x)j M for all x 2 D.
In the example we have shown that x29x+1 is bounded when considered as a function on
D = fx j 1 x 5g. On the other hand, if we consider the same polynomial as a function on the
whole real line R, then it is not bounded.
If a function f : D!R is bounded, then we can talk about its supremum and its infimum. We
write
sup
x2D
f (x) := sup f (D);
inf
x2D
f (x) := inf f (D):
To illustrate some common issues, let us prove the following proposition.
Proposition 1.3.7. If f : D!R and g: D!R are bounded functions and
f (x) g(x) for all x 2 D;
then
sup
x2D
f (x) sup
x2D
g(x) and inf
x2D
f (x) inf
x2D
g(x): (1.1)
You should be careful with the variables. The x on the left side of the inequality in (1.1) is
different from the x on the right. You should really think of the first inequality as
sup
x2D
f (x) sup
y2D
g(y):
Let us prove this inequality. If b is an upper bound for g(D), then f (x) g(x) b and hence b is
an upper bound for f (D). Therefore taking the least upper bound we get that for all x
f (x) sup
y2D
g(y):
But that means that supy2D g(y) is an upper bound for f (D), hence is greater than or equal to the
least upper bound of f (D).
sup
x2D
f (x) sup
y2D
g(y):
The second inequality (the statement about the inf) is left as an exercise.
Do note that a common mistake is to conclude that
sup
x2D
f (x) inf
y2D
g(y): (1.2)
The inequality (1.2) is not true given the hypothesis of the claim above. For this stronger inequality
you need the stronger hypothesis
f (x) g(y) for all x 2 D and y 2 D.
The proof is left as an exercise.
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