2.3 Real Numbers
The union of the set of rational numbers with the set of irrational numbers is the set of real numbers, denoted \(\mathbb {R}\). Geometrically \(\mathbb {R}\) is the number line, with every point on the line corresponding to exactly one real number and no gaps left over.
Subsets of \(\mathbb {R}\) that consist of everything between two endpoints are written with interval notation, using brackets of two kinds: \[(\cdot ,\cdot ),\quad (\cdot ,\cdot ],\quad [\cdot ,\cdot ),\quad [\cdot ,\cdot ].\] A round bracket, \((\) or \()\), means the endpoint is not included; a square bracket, \([\) or \(]\), means it is. The two can be mixed in one interval, and an interval with both endpoints excluded is called open, with both included closed.
- \([-2,0)\). This means \(-2\) is part of the set while 0 is not part of the set.
- Set builder notation can also be used to describe the set \[[-2,0)=\{x\in \mathbb {R}:-2\leq x<0\}\]
- When using interval notation the smaller number must be on the left while the bigger one must be on right side.
- \(\infty \) is not a real number; it is a symbol recording that an interval continues without end. The bracket next to it is therefore always round: \([2,\infty )\) is correct and \([2,\infty ]\) is not, because there is no endpoint there to include.
Note 2.16. The four brackets matter more than they look. \([0,1]\) has a largest element and \([0,1)\) does not, and that single difference decides whether a function on the interval is guaranteed to attain a maximum. Being careful with brackets now saves confusion later.
Operations on \(\mathbb {R}\)
Let \(A=\{x\in \mathbb {R}:-7\leq x<3\}\) and \(B=\{x\in \mathbb {R}:x\geq -1\}\)
Find
- 1.
- \(A\cap B\)
- 2.
- \(A'\)
Solution.
- 1.
- In interval notation \(A=[-7,3)\) and \(B=[-1,\infty )\), since \(B\) has no upper bound. Their intersection needs a point in both, so it starts where the later interval starts and ends where the earlier one ends: \[A\cap B=[-1,3).\]
- 2.
- Taking \(\mathbb {R}\) as the universal set, \(A'\) is everything outside \([-7,3)\): \[A'=(-\infty ,-7)\cup [3,\infty ).\] Note how the brackets swap. \(-7\) belonged to \(A\), so it is excluded from \(A'\); \(3\) did not belong to \(A\), so it is included in \(A'\). A complement turns each included endpoint into an excluded one and the other way round.
Let \(E=(-7,10]\) be the universal set and let \(A=[-3,7]\), \(B=(-2,5)\) and \(C=(-6,10]\).
Find
- 1.
- \(A\cap B\)
- 2.
- \(E-C\)
Solution.
- 1.
- \(A=[-3,7]\) and \(B=(-2,5)\), and \(B\) is entirely inside \(A\), so \(A\cap B=(-2,5)=B\).
- 2.
- \(E-C\) is what remains of \(E=(-7,10]\) after removing \(C=(-6,10]\), namely \[E-C=(-7,-6].\] The left bracket stays round because \(-7\) was never in \(E\) to begin with, so it cannot appear in a subset of \(E\). The right bracket is square because \(-6\) is in \(E\) and is not in \(C\), which starts just above it.
Properties Of \(\mathbb {R}\)
Real numbers have certain properties which allow us to compare any two real numbers or to perform
some algebraic computation.
- 1.
- Algebraic Properties
If \(x\), \(y\) and \(z\) are any three numbers, then- (a)
- \(x+y=y+x\) Commutative law
- (b)
- \( \begin {matrix} x+(y+z) & = & (x+y)+z\\ x(yz) & = & (xy)z\\ \end {matrix} \) Associative Law
- (c)
- \(x+0=0+x=x,\quad \) 0 is addictive identity
- (d)
- \(x\cdot 1=1\cdot x=x,\quad \) 1 is multiplicative identity
- (e)
- \(x+(-x)=x-x=-x+x=0\)
- (f)
- \(x(y+z)=xy+xz\), Distributive Law
- 2.
- Order Relations:
If \(x\) and \(y\) are any two real numbers then- (a)
- \(x\) is greater than \(y\) written \(x>y\) if \(x-y>0.\)
- (b)
- \(x\) is less than \(y\) written \(x<y\) if \(x-y<0\).
- (c)
- \(x\) is equal to \(y\) written \(x=y\) if \(x-y=0\).
- 3.
- Absolute Value:
Definition 2.19. The absolute value of a real number \(x\), written \(\left |x\right |\), is \[\left |x\right |= \begin {cases} x, & \text {if}\quad x\geq 0\\ -x, & \text {if} \quad x<0 . \end {cases}\]
The absolute value of a real number is never negative. It is not always positive: \(\left |0\right |=0\), and zero is not positive. The correct word is non-negative, and the distinction matters whenever an argument needs \(\left |x\right |>0\), which requires \(x\neq 0\) as well.
For example
- (a).
- \(\left |8\right |=8\);
- (b).
- \(\left |-23\right |=-(-23)=23\).
Note 2.20. The definition is easy to misread as “drop the minus sign”. Read it geometrically instead: \(\left |x\right |\) is the distance from \(x\) to \(0\) on the number line, and distances carry no sign. More useful still, \[\left |x-a\right | \text { is the distance between } x \text { and } a .\] Read that way several things become obvious rather than needing algebra. \(\left |x-7\right |=11\) asks which points sit \(11\) away from \(7\) — plainly \(18\) and \(-4\), one on each side. \(\left |x\right |<k\) asks which points lie less than \(k\) from the origin, which is the stretch between \(-k\) and \(k\). Every absolute-value equation and inequality in this course can be checked against that picture.
Note also that \(-x\) is not “a negative number”. If \(x=-5\) then \(-x=5\), which is precisely why the second branch returns a non-negative answer.
Find the value of \(|a-b|\) when
- (i).
- \(a=7,\quad b=4\)
- (ii).
- \(a=-6,\quad b=8\)
- (iii).
- \(a=2,\quad b=11\)
Solution. \(\left |a-b\right |\) is the distance between \(a\) and \(b\), so the order makes no difference and the answer is never negative.
- (i).
- \(\left |7-4\right |=\left |3\right |=3\).
- (ii).
- \(\left |-6-8\right |=\left |-14\right |=14\).
- (iii).
- \(\left |2-11\right |=\left |-9\right |=9\).
In (ii) and (iii) the quantity inside the bars is negative and the second branch applies: \(\left |-14\right |=-(-14)=14\).
Given that \(a<b\). Express \(P\) in terms of \(a\) and \(b\), if \(\frac {|P-a|}{|b-a|}=\frac {3}{4}\).
Solution.
\(\frac {|P-a|}{|b-a|}=\frac {3}{4}\quad \implies \quad \frac {|P-a|}{b-a}=\frac {3}{4}\)
\(\implies \quad |P-a|=\frac {3}{4}(b-a)\)
\(\implies \quad |P-a|=\frac {3}{4}b-\frac {3}{4}a\)
case I: if \(P\geq a\) then \(P-a\geq 0\) \begin {align*} \implies \quad P-a & = \frac {3b}{4}-\frac {3a}{4}\quad \implies \quad P = \frac {3b}{4}-\frac {3a}{4}+a\\\\ & = \frac {3b}{4}+\frac {-3a+4a}{4}\\\\ \implies \quad P & = \frac {3b}{4}+\frac {a}{4}\\\\ \end {align*}
Case II: If \(P<a\) then \(P-a<0\).
Therefore, \(|P-a|=-(P-a)=-P+a\) in this case \begin {align*} -P+a & = \frac {3b}{4}-\frac {3a}{4}\quad \implies \quad -P=\frac {3b}{4}-\frac {3a}{4}-a\\\\ - P & = \frac {3b}{4}-\frac {7a}{4}\\\\ \implies \quad P & = \frac {7a}{4}-\frac {3b}{4}\\\\ \end {align*}
Solve the equation, \(|x-7|=11\).
Case I: \(x\geq 7\), so that \(x-7\geq 0\) and \(\left |x-7\right |=x-7\). Then \[x-7=11\quad \implies \quad x=7+11=18 .\]
Case II: \(x<7\), so that \(x-7<0\) and \(\left |x-7\right |=-(x-7)\). Then \[-(x-7)=11\quad \implies \quad x-7=-11 \quad \implies \quad x=-11+7=-4 .\]
Both values satisfy the original equation, so \(x=18\) or \(x=-4\) — the two points lying \(11\) units from \(7\), one on
each side, exactly as the distance reading predicts.
- Let \(k\) be a positive real number consider the inequality \(|x|<k\).
If \(x\geq 0\) then \(|x|=x\) so that \(|x|<k\quad \implies \quad x<k\, \cdots \,(1)\)
If \(x<0\) then \(|x|=-x\) so that \(|x|<k\quad \implies \quad -x<k\quad \\ \implies \quad -k<x\, \cdots \, (2)\)
Now, combining the two inequalities we conclude that \(|x|<k \iff -k<x<k\).
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