THE REAL NUMBERS
Section 2.2 Absolute Value and the Real Line
fromjbj ¼ jbaþaj jbaj þ jaj, we obtainjabj ¼ jbaj jaj jbj. If we combine these two inequalities, using 2.2.2(c), we get the inequality in (a).
(b) Replacebin the Triangle Inequality bybto getjabj jaj þ j bj. Sincej bj ¼
jbjwe obtain the inequality in (b). Q.E.D.
A straightforward application of Mathematical Induction extends the Triangle In- equality to any finite number of elements ofR.
2.2.5 Corollary If a1;a2;. . .;an are any real numbers, then ja1þa2þ þanj ja1j þ ja2j þ þ janj:
The following examples illustrate how the properties of absolute value can be used.
2.2.6 Examples (a) Determine the set Aofx2R such thatj2xþ3j<7.
From a modification of 2.2.2(c) for the case of strict inequality, we see thatx2Aif and only if7<2xþ3<7, which is satisfied if and only if10<2x<4. Dividing by 2, we conclude that A¼ fx2R:5<x<2g.
(b) Determine the set B:¼ fx2R:jx1j<jxjg.
One method is to consider cases so that the absolute value symbols can be removed.
Here we take the cases
ðiÞx1; ðiiÞ0x<1; ðiiiÞx<0:
(Why did we choose these three cases?) In case (i) the inequality becomesx1<x, which is satisfied without further restriction. Therefore allxsuch thatx1 belong to the setB. In case (ii), the inequality becomesðx1Þ<x, which requires thatx>12. Thus, this case contributes all x such that 12<x<1 to the set B. In case (iii), the inequality becomesðx1Þ<x, which is equivalent to 1<0. Since this statement is false, no value ofxfrom case (iii) satisfies the inequality. Forming the union of the three cases, we conclude thatB¼x2R:x>12
.
There is a second method of determining the setBbased on the fact thata<bif and only ifa2<b2when botha0 andb0. (See 2.1.13(a).) Thus, the inequalityjx1j<
jxjis equivalent to the inequalityjx1j2 <jxj2. Sincejaj2¼a2for anyaby 2.2.2(b), we can expand the square to obtainx22xþ1<x2, which simplifies to x>12. Thus, we again find thatB¼x2R:x>12
. This method of squaring can sometimes be used to advantage, but often a case analysis cannot be avoided when dealing with absolute values.
A graphical view of the inequality is obtained by sketching the graphs ofy¼ jxjand y¼ jx1j, and interpreting the inequalityjx1j<jxj to mean that the graph ofy¼ jx1jlies underneath the graph of y¼ jxj. See Figure 2.2.1.
y
x y = | x |
y = | x – 1|
1
1 2
Figure 2.2.1 jx1j<jxj
2.2 ABSOLUTE VALUE AND THE REAL LINE 33
(c) Solve the inequalityj2x1j xþ1.
There are two cases to consider. Ifx12, thenj2x1j ¼ 2xþ1 and the inequality becomes2xþ1xþ1, which isx0. Thus case one gives us 0x12. For case two we assumex12, which gives us the inequality 2x1xþ1, orx2. Sincex12, we obtain12x2. Combining the two cases, we get 0x2. See Figure 2.2.2.
(d) Let the function f be defined by fðxÞ:¼ ð2x2þ3xþ1Þ=ð2x1Þ for 2x<3.
Find a constantM such thatjfðxÞj M for allxsatisfying 2x3.
We consider separately the numerator and denominator of jfðxÞj ¼j2x2þ3xþ1j
j2x1j : From the Triangle Inequality, we obtain
j2x2þ3xþ1j 2jxj2þ3jxj þ1232þ33þ1¼28
sincejxj 3 for thex under consideration. Also,j2x1j 2jxj 1221¼3 sincejxj 2 for thexunder consideration. Thus, 1=j2x1j 1=3 forx2. (Why?) Therefore, for 2x3 we havejfðxÞj 28=3. Hence we can takeM ¼28=3. (Note that we have found one such constant M; evidently any numberH>28=3 will also satisfy jfðxÞj H. It is also possible that 28=3 is not the smallest possible choice
for M.) &
The Real Line
A convenient and familiar geometric interpretation of the real number system is the real line. In this interpretation, the absolute valuejajof an elementainR is regarded as the distance fromato the origin 0. More generally, thedistancebetween elementsaandbinR isjabj. (See Figure 2.2.3.)
y
x y = x + 1
2 1 –1
y = |2x – 1| (0, 1)
(2, 3)
Figure 2.2.2 j2x1j xþ1
Figure 2.2.3 The distance betweena¼ 2 andb¼3
Later we will need precise language to discuss the notion of one real number being
‘‘close to’’ another. Ifais a given real number, then saying that a real numberxis ‘‘close to’’ashould mean that the distancejxajbetween them is ‘‘small.’’ A context in which this idea can be discussed is provided by the terminology of neighborhoods, which we now define.
2.2.7 Definition Let a2R and e>0. Then the e-neighborhood of a is the set VeðaÞ:¼ fx2R:jxaj<eg.
For a2R, the statement that x belongs to VeðaÞ is equivalent to either of the statements (see Figure 2.2.4)
e<xa<e () ae<x<aþe:
2.2.8 Theorem Let a2R.If x belongs to the neighborhood VeðaÞfor everye>0, then x¼a.
Proof. If a particularxsatisfiesjxaj<efor everye>0, then it follows from 2.1.9 that
jxaj ¼0, and hencex¼a. Q.E.D.
2.2.9 Examples (a) LetU:¼ fx:0<x<1g. Ifa2U, then letebe the smaller of the two numbersaand 1a. Then it is an exercise to show thatVeðaÞis contained inU. Thus each element ofUhas somee-neighborhood of it contained inU.
(b) IfI:¼ fx:0x1g, then for anye>0, thee-neighborhoodVeð0Þof 0 contains points not inI, and soVeð0Þis not contained inI. For example, the numberxe:¼ e=2 is in Veð0Þbut not inI.
(c) Ifjxaj<eandjybj<e, then the Triangle Inequality implies that jðxþyÞ ðaþbÞj ¼ jðxaÞ þ ðybÞj
jxaj þ jybj<2e:
Thus ifx,ybelong to thee-neighborhoods ofa,b, respectively, thenxþybelongs to the 2e-neighborhood ofaþb(but not necessarily to thee-neighborhood ofaþb). &
Exercises for Section 2.2
1. Ifa;b2R andb6¼0, show that:
(a) jaj ¼ ffiffiffiffiffi a2
p ; (b) ja=bj ¼ jaj=jbj:
2. Ifa;b2R, show thatjaþbj ¼ jaj þ jbjif and only ifab0.
3. Ifx;y;z2Randxz, show thatxyzif and only ifjxyj þ jyzj ¼ jxzj. Interpret this geometrically.
Figure 2.2.4 Ane-neighborhood ofa
2.2 ABSOLUTE VALUE AND THE REAL LINE 35
4. Show thatjxaj<eif and only ifae<x<aþe.
5. Ifa<x<banda<y<b, show thatjxyj<ba. Interpret this geometrically.
6. Find allx2R that satisfy the following inequalities:
(a) j4x5j 13; (b) jx21j 3:
7. Find allx2R that satisfy the equationjxþ1j þ jx2j ¼7.
8. Find all values ofxthat satisfy the following equations:
(a) xþ1¼ j2x1j, (b) 2x1¼ jx5j.
9. Find all values ofxthat satisfy the following inequalities. Sketch graphs.
(a) jx2j xþ1, (b) 3jxj 2x:
10. Find allx2R that satisfy the following inequalities.
(a) jx1j>jxþ1j; (b) jxj þ jxþ1j<2:
11. Sketch the graph of the equationy¼ jxj jx1j.
12. Find allx2R that satisfy the inequality 4<jxþ2j þ jx1j<5.
13. Find allx2R that satisfy bothj2x3j<5 andjxþ1j>2 simultaneously.
14. Determine and sketch the set of pairsðx;yÞinRRthat satisfy:
(a) jxj ¼ jyj; (b) jxj þ jyj ¼1;
(c) jxyj ¼2, (d) jxj jyj ¼2: 15. Determine and sketch the set of pairs (x,y) inRR that satisfy:
(a) jxj jyj; (b) jxj þ jyj 1;
(c) jxyj 2; (d) jxj jyj 2:
16. Lete>0 andd>0, anda2R. Show thatVeðaÞ \VdðaÞandVeðaÞ [VdðaÞareg-neighbor- hoods ofafor appropriate values ofg.
17. Show that ifa;b2R, anda6¼b, then there existe-neighborhoodsUofaandVofbsuch that U\V¼ ;.
18. Show that ifa;b2R then
(a) maxfa;bg ¼12ðaþbþ jabjÞ and minfa;bg ¼12ðaþb jabjÞ:
(b) minfa;b;cg ¼minfminfa;bg;cg:
19. Show that if a;b;c2R, then the ‘‘middle number’’ is midfa;b;cg ¼ minfmaxfa;bg;
maxfb;cg;maxfc;agg.