Link to the curriculum
The set of real numbers, absolute value (definition, properties, interpretation as distance), intervals and inequalities in $\mathbb{R}$ are skills from grades 7-8, which we reactivate at the start of grade 9.
The integer part and the fractional part are new notions. The grade 9 curriculum introduces them as examples of functions (the absolute value function, the integer part function, the fractional part function) in Learning Unit 4 - Functions. General properties. Here we work with them as numbers; their graphs and properties as functions are studied in Unit 4.
1. Sets of numbers
- ${\mathbb{N}} = \{ 0,\ 1,\ 2,\ 3,\ \ldots\}$ - the natural numbers; ${\mathbb{Z}} = \{\ldots,\ - 2,\ - 1,\ 0,\ 1,\ 2,\ \ldots\}$ - the integers;
- ${\mathbb{Q}} = \{\frac{a}{b}\text{ | }a \in {\mathbb{Z}},\ b \in {\mathbb{Z}}^{*}\}$ - the rational numbers (finite or repeating decimal expansion);
- ${\mathbb{R}} \smallsetminus {\mathbb{Q}}$ - the irrational numbers (infinite, non-repeating decimal expansion), for example $\sqrt{2}$, $\sqrt{3}$, $\pi$;
- ${\mathbb{N}} \subset {\mathbb{Z}} \subset {\mathbb{Q}} \subset {\mathbb{R}}$; notation: ${\mathbb{R}}^{*} = {\mathbb{R}} \smallsetminus \{ 0\}$, ${\mathbb{R}}_{+} = \lbrack 0,\ + \infty)$, ${\mathbb{R}}_{-} = ( - \infty,\ 0\rbrack$.
2. Order relation on ℝ. Intervals
2.1. Properties of inequalities
For any $a,\ b,\ c,\ d \in {\mathbb{R}}$:
- $a \leq b$ and $b \leq c$ $\Rightarrow$ $a \leq c$ (transitivity); $a \leq b$ $\Rightarrow$ $a + c \leq b + c$;
- $a \leq b$ and $c \leq d$ $\Rightarrow$ $a + c \leq b + d$ (inequalities pointing the same way can be added);
- $a \leq b$ and $c > 0$ $\Rightarrow$ $ac \leq bc$; $a \leq b$ and $c < 0$ $\Rightarrow$ $ac \geq bc$ (the inequality reverses!);
- $0 < a \leq b$ $\Rightarrow$ $\frac{1}{a} \geq \frac{1}{b}$; $x^{2} \geq 0$ for any $x \in {\mathbb{R}}$, and $x^{2} = 0 \Leftrightarrow x = 0$.
2.2. Intervals of real numbers
Let $a,\ b \in {\mathbb{R}}$, $a < b$.
| Type of interval | Notation | Set |
|---|---|---|
| closed | $$\lbrack a,\ b\rbrack$$ | $$\{ x \in {\mathbb{R}}\text{ | }a \leq x \leq b\}$$ |
| open | $$(a,\ b)$$ | $$\{ x \in {\mathbb{R}}\text{ | }a < x < b\}$$ |
| half-open |
$$\lbrack a,\ b)$$ $$(a,\ b\rbrack$$ |
$$\{ x \in {\mathbb{R}}\text{ | }a \leq x < b\}$$ $$\{ x \in {\mathbb{R}}\text{ | }a < x \leq b\}$$ |
| unbounded |
$\lbrack a,\ + \infty)$, $(a,\ + \infty)$ $( - \infty,\ b\rbrack$, $( - \infty,\ b)$ |
$\{ x \in {\mathbb{R}}\text{ | }x \geq a\}$, $\{ x \in {\mathbb{R}}\text{ | }x > a\}$ $\{ x \in {\mathbb{R}}\text{ | }x \leq b\}$, $\{ x \in {\mathbb{R}}\text{ | }x < b\}$ |
Example
For $A = \lbrack - 2,\ 3)$ and $B = (1,\ 5\rbrack$: $A \cap B = (1,\ 3)$, $A \cup B = \lbrack - 2,\ 5\rbrack$, $A \smallsetminus B = \lbrack - 2,\ 1\rbrack$, $B \smallsetminus A = \lbrack 3,\ 5\rbrack$.
At $\pm \infty$ the interval is always open: we write $\lbrack a,\ + \infty)$, never $\lbrack a,\ + \infty\rbrack$.
3. Absolute value of a real number
3.1. Definition. Geometric interpretation
Definition
The absolute value of the real number $x$ is:
$$|x| = \left\{ \begin{matrix} x,\ & \text{if }x \geq 0 \\ - x,\ & \text{if }x < 0 \end{matrix} \right.$$
- $|x|$ is the distance from the point with coordinate $x$ to the origin of the axis; $|x - a|$ is the distance between the points with coordinates $x$ and $a$.
- Examples: $| - 5| = 5$; $|0| = 0$; $\left| \sqrt{2} - 2 \right| = 2 - \sqrt{2}$ (because $\sqrt{2} < 2$); $|\pi - 3| = \pi - 3$ (because $\pi > 3$).
3.2. Properties of absolute value
For any $x,\ y \in {\mathbb{R}}$ and $a > 0$:
| No. | Property | Remarks / examples |
|---|---|---|
| 1 | $|x| \geq 0$; $|x| = 0 \Leftrightarrow x = 0$ | $\left| E(x) \right| = - 3$ has no solutions |
| 2 | $| - x| = |x|$; $- |x| \leq x \leq |x|$ | $$|3 - x| = |x - 3|$$ |
| 3 | $|x \cdot y| = |x| \cdot |y|$; $\left| \frac{x}{y} \right| = \frac{|x|}{|y|}$, $y \neq 0$ | $$| - 2x| = 2|x|$$ |
| 4 | $|x|^{2} = x^{2}$; $\sqrt{x^{2}} = |x|$ | $$\sqrt{\left( 1 - \sqrt{3} \right)^{2}} = \sqrt{3} - 1$$ |
| 5 | $|x + y| \leq |x| + |y|$ (the triangle inequality) | equality $\Leftrightarrow$ $xy \geq 0$ |
| 6 | $$\left| |x| - |y| \right| \leq |x - y|$$ | follows from property 5 |
| 7 | $|x| = |y|$ $\Leftrightarrow$ $x = y$ or $x = - y$ | $|x| = |y|$ $\Leftrightarrow$ $x^{2} = y^{2}$ |
| 8 | $|x| \leq a$ $\Leftrightarrow$ $- a \leq x \leq a$ $\Leftrightarrow$ $x \in \lbrack - a,\ a\rbrack$ | $|x| < a$ $\Leftrightarrow$ $x \in ( - a,\ a)$ |
| 9 | $|x| \geq a$ $\Leftrightarrow$ $x \leq - a$ or $x \geq a$ | $$x \in ( - \infty,\ - a\rbrack \cup \lbrack a,\ + \infty)$$ |
3.3. Calculations: removing the absolute value
We determine the sign of the expression inside the absolute value (by comparing the numbers), then apply the definition.
Example 1
$a = \left| \sqrt{3} - 2 \right| + \left| 1 - \sqrt{3} \right|$. Since $\sqrt{3} < 2$ and $\sqrt{3} > 1$: $a = \left( 2 - \sqrt{3} \right) + \left( \sqrt{3} - 1 \right) = 1 \in {\mathbb{N}}$.
Example 2
For $x \in \lbrack - 2,\ 1\rbrack$: $x - 1 \leq 0$ and $x + 2 \geq 0$, so $|x - 1| + |x + 2| = (1 - x) + (x + 2) = 3$ (it does not depend on $x$).
Example 3
$\sqrt{\left( \sqrt{5} - 3 \right)^{2}} + \sqrt{\left( \sqrt{5} - 2 \right)^{2}} = \left| \sqrt{5} - 3 \right| + \left| \sqrt{5} - 2 \right| = 3 - \sqrt{5} + \sqrt{5} - 2 = 1$.
3.4. Equations with absolute value
| Type of equation | Method of solving |
|---|---|
| $$\left| E(x) \right| = a$$ |
$a < 0$: $S = \varnothing$; $a = 0$: $E(x) = 0$; $a > 0$: $E(x) = a$ or $E(x) = - a$. |
| $$\left| E(x) \right| = \left| F(x) \right|$$ | $E(x) = F(x)$ or $E(x) = - F(x)$. |
| $$\left| E(x) \right| = F(x)$$ |
condition: $F(x) \geq 0$; $E(x) = F(x)$ or $E(x) = - F(x)$; we keep only the solutions that satisfy the condition. |
| sums of absolute values | the interval method: we split the number line at the zeros of the expressions inside the absolute values and solve on each interval. |
Example 4
$|2x - 1| = 5$ $\Leftrightarrow$ $2x - 1 = 5$ or $2x - 1 = - 5$ $\Leftrightarrow$ $x = 3$ or $x = - 2$. So $S = \{ - 2,\ 3\}$.
Example 5
$|x + 1| = 2x - 4$. Condition: $2x - 4 \geq 0 \Leftrightarrow x \geq 2$.
$x + 1 = 2x - 4 \Rightarrow x = 5$ (valid); $x + 1 = - (2x - 4) \Rightarrow x = 1$ (not valid, $1 < 2$). So $S = \{ 5\}$.
Example 6
$|x| + |x - 2| = 4$. The zeros of the expressions inside the absolute values are $0$ and $2$:
| $$x$$ | $$( - \infty,\ 0)$$ | $$\lbrack 0,\ 2)$$ | $$\lbrack 2,\ + \infty)$$ |
|---|---|---|---|
| $$|x|$$ | $$- x$$ | $$x$$ | $$x$$ |
| $$|x - 2|$$ | $$2 - x$$ | $$2 - x$$ | $$x - 2$$ |
| the equation becomes | $2 - 2x = 4 \Rightarrow x = - 1$ ✓ | $2 = 4$, false | $2x - 2 = 4 \Rightarrow x = 3$ ✓ |
The solutions found belong to their corresponding intervals, so $S = \{ - 1,\ 3\}$.
3.5. Inequalities with absolute value
| Type of inequality | Equivalent to |
|---|---|
| $\left| E(x) \right| \leq a$, $a > 0$ | $$- a \leq E(x) \leq a$$ |
| $\left| E(x) \right| \geq a$, $a > 0$ | $E(x) \leq - a$ or $E(x) \geq a$ (union of intervals) |
| $\left| E(x) \right| < a$, $a \leq 0$ | $$S = \varnothing$$ |
| $\left| E(x) \right| \geq a$, $a \leq 0$ | $$S = {\mathbb{R}}$$ |
Example 7
$|3x - 2| < 4$ $\Leftrightarrow$ $- 4 < 3x - 2 < 4$ $\Leftrightarrow$ $- 2 < 3x < 6$ $\Leftrightarrow$ $- \frac{2}{3} < x < 2$. So $S = \left( - \frac{2}{3},\ 2 \right)$.
Example 8
$|x + 3| \geq 2$ $\Leftrightarrow$ $x + 3 \leq - 2$ or $x + 3 \geq 2$ $\Leftrightarrow$ $x \leq - 5$ or $x \geq - 1$. So $S = ( - \infty,\ - 5\rbrack \cup \lbrack - 1,\ + \infty)$.
Example 9 (distance)
$|x - 4| \leq 1$: the points at distance at most $1$ from the point $4$, that is $S = \lbrack 3,\ 5\rbrack$.
4. Integer part and fractional part of a real number
4.1. Definitions
Definition
For any $x \in {\mathbb{R}}$ there is a single integer $k$ such that $k \leq x < k + 1$. This number is called the integer part of $x$ and is denoted $\lbrack x\rbrack$. So $\lbrack x\rbrack$ is the largest integer less than or equal to $x$.
The fractional part of $x$ is $\left\{ x \right\} = x - \lbrack x\rbrack$. (Some books use the notation $\lfloor x\rfloor$.)
| $$x$$ | $$\lbrack x\rbrack$$ | $$\left\{ x \right\}$$ | Justification |
|---|---|---|---|
| $$\text{2,6}$$ | $$2$$ | $$\text{0,6}$$ | $$2 \leq \text{2,6} < 3$$ |
| $$- \text{2,6}$$ | $$- 3$$ | $$\text{0,4}$$ | $$- 3 \leq - \text{2,6} < - 2$$ |
| $$5$$ | $$5$$ | $$0$$ | $$5 \in {\mathbb{Z}}$$ |
| $$- \text{0,25}$$ | $$- 1$$ | $$\text{0,75}$$ | $$- 1 \leq - \text{0,25} < 0$$ |
| $$\sqrt{10}$$ | $$3$$ | $$\sqrt{10} - 3$$ | $$3 < \sqrt{10} < 4$$ |
| $$- \sqrt{2}$$ | $$- 2$$ | $$2 - \sqrt{2}$$ | $$- 2 < - \sqrt{2} < - 1$$ |
Careful!
$\left\lbrack - \text{2,6} \right\rbrack = - 3$, not $- 2$. For negative numbers, the integer part is not obtained by "cutting off the decimals". The fractional part is always in $\lbrack 0,\ 1)$: $\left\{ - \text{2,6} \right\} = - \text{2,6} - ( - 3) = \text{0,4}$.
4.2. Properties
For any $x,\ y \in {\mathbb{R}}$ and $k \in {\mathbb{Z}}$:
| No. | Property | Remarks / examples |
|---|---|---|
| 1 | $\lbrack x\rbrack \leq x < \lbrack x\rbrack + 1$; equivalently: $x - 1 < \lbrack x\rbrack \leq x$ | $$\lbrack x\rbrack \in {\mathbb{Z}}$$ |
| 2 | $\lbrack x\rbrack = k$ $\Leftrightarrow$ $k \leq x < k + 1$ $\Leftrightarrow$ $x \in \lbrack k,\ k + 1)$ | $\lbrack x\rbrack = \frac{5}{2}$ has no solutions |
| 3 | $x = \lbrack x\rbrack + \left\{ x \right\}$; $0 \leq \left\{ x \right\} < 1$; $\left\{ x \right\} = 0 \Leftrightarrow x \in {\mathbb{Z}}$ | $\left\{ x \right\} = \text{1,2}$ has no solutions |
| 4 | $\lbrack x + k\rbrack = \lbrack x\rbrack + k$; $\left\{ x + k \right\} = \left\{ x \right\}$ | $$\lbrack x + 3\rbrack = \lbrack x\rbrack + 3$$ |
| 5 |
$\lbrack x\rbrack + \lbrack - x\rbrack = 0$ if $x \in {\mathbb{Z}}$, $= - 1$ if $x \notin {\mathbb{Z}}$ $\left\{ x \right\} + \left\{ - x \right\} = 0$ if $x \in {\mathbb{Z}}$, $= 1$ if $x \notin {\mathbb{Z}}$ |
$$\lbrack\pi\rbrack + \lbrack - \pi\rbrack = 3 + ( - 4) = - 1$$ |
| 6 | $x \leq y$ $\Rightarrow$ $\lbrack x\rbrack \leq \lbrack y\rbrack$ | the converse implication is false |
| 7 | $$\lbrack x\rbrack + \lbrack y\rbrack \leq \lbrack x + y\rbrack \leq \lbrack x\rbrack + \lbrack y\rbrack + 1$$ | equality on the left $\Leftrightarrow$ $\left\{ x \right\} + \left\{ y \right\} < 1$ |
| 8 | $\lbrack x\rbrack \leq k \Leftrightarrow x < k + 1$; $\lbrack x\rbrack < k \Leftrightarrow x < k$; $\lbrack x\rbrack \geq k \Leftrightarrow x \geq k$ | useful for inequalities |
| 9 | $\lbrack x\rbrack + \left\lbrack x + \frac{1}{2} \right\rbrack = \lbrack 2x\rbrack$ (Hermite's identity) | extension - proof: worksheet, item 29 |
4.3. Calculations
Example 10
$49 \leq 50 < 64$ $\Rightarrow$ $7 \leq \sqrt{50} < 8$, so $\left\lbrack \sqrt{50} \right\rbrack = 7$, $\left\{ \sqrt{50} \right\} = \sqrt{50} - 7 = 5\sqrt{2} - 7$, and $\left\lbrack - \sqrt{50} \right\rbrack = - 8$.
Example 11
$S = \left\lbrack \sqrt{1} \right\rbrack + \left\lbrack \sqrt{2} \right\rbrack + \ldots + \left\lbrack \sqrt{8} \right\rbrack$. For $n \in \{ 1,\ 2,\ 3\}$ we have $\left\lbrack \sqrt{n} \right\rbrack = 1$, and for $n \in \{ 4,\ \ldots,\ 8\}$ we have $\left\lbrack \sqrt{n} \right\rbrack = 2$. So $S = 3 \cdot 1 + 5 \cdot 2 = 13$.
4.4. Equations with integer part and fractional part
| Type of equation | Method of solving |
|---|---|
| $\left\lbrack E(x) \right\rbrack = k$, $k \in {\mathbb{Z}}$ | $k \leq E(x) < k + 1$ (if the right-hand side is not an integer, $S = \varnothing$). |
| $$\left\lbrack E(x) \right\rbrack = F(x)$$ | we denote $F(x) = k \in {\mathbb{Z}}$, express $x$ in terms of $k$, impose $k \leq E(x) < k + 1$, find the integer values of $k$, then $x$. |
| equations with $\lbrack x\rbrack$ and $\left\{ x \right\}$ | we write $x = k + f$, with $k = \lbrack x\rbrack \in {\mathbb{Z}}$ and $f = \left\{ x \right\} \in \lbrack 0,\ 1)$. |
| expressions in $\lbrack x\rbrack$ | the substitution $t = \lbrack x\rbrack$, with $t \in {\mathbb{Z}}$. |
Example 12
$\left\lbrack \frac{x - 1}{2} \right\rbrack = 3$ $\Leftrightarrow$ $3 \leq \frac{x - 1}{2} < 4$ $\Leftrightarrow$ $6 \leq x - 1 < 8$ $\Leftrightarrow$ $7 \leq x < 9$. So $S = \lbrack 7,\ 9)$.
Example 13
$\left\lbrack \frac{x + 2}{3} \right\rbrack = \frac{x - 1}{2}$. We denote $\frac{x - 1}{2} = k \in {\mathbb{Z}}$, so $x = 2k + 1$ and $\frac{x + 2}{3} = \frac{2k + 3}{3}$.
$k \leq \frac{2k + 3}{3} < k + 1$ $\Leftrightarrow$ $3k \leq 2k + 3 < 3k + 3$ $\Leftrightarrow$ $0 < k \leq 3$ $\Rightarrow$ $k \in \{ 1,\ 2,\ 3\}$ $\Rightarrow$ $S = \{ 3,\ 5,\ 7\}$.
Example 14
$2\lbrack x\rbrack = 3\left\{ x \right\}$. With $x = k + f$: $2k = 3f$, so $f = \frac{2k}{3} \in \lbrack 0,\ 1)$ $\Leftrightarrow$ $0 \leq 2k < 3$ $\Rightarrow$ $k \in \{ 0,\ 1\}$.
$k = 0 \Rightarrow f = 0 \Rightarrow x = 0$; $k = 1 \Rightarrow f = \frac{2}{3} \Rightarrow x = \frac{5}{3}$. So $S = \{ 0,\ \frac{5}{3}\}$.
Example 15
$\lbrack x\rbrack^{2} - 3\lbrack x\rbrack + 2 = 0$. With $t = \lbrack x\rbrack$: $(t - 1)(t - 2) = 0$ $\Rightarrow$ $\lbrack x\rbrack \in \{ 1,\ 2\}$ $\Rightarrow$ $x \in \lbrack 1,\ 2) \cup \lbrack 2,\ 3) = \lbrack 1,\ 3)$.
4.5. Inequalities with integer part
Example 16
$\lbrack 2x - 1\rbrack < 3$ $\Leftrightarrow$ $\lbrack 2x - 1\rbrack \leq 2$ $\Leftrightarrow$ $2x - 1 < 3$ $\Leftrightarrow$ $x < 2$. So $S = ( - \infty,\ 2)$.
Example 17
$- 1 \leq \left\lbrack \frac{x}{2} \right\rbrack \leq 1$ $\Leftrightarrow$ $- 1 \leq \frac{x}{2} < 2$ $\Leftrightarrow$ $- 2 \leq x < 4$. So $S = \lbrack - 2,\ 4)$.
5. Common mistakes
| Wrong | Correct |
|---|---|
| $$\sqrt{x^{2}} = x$$ | $$\sqrt{x^{2}} = |x|$$ |
| $$|a + b| = |a| + |b|$$ | $|a + b| \leq |a| + |b|$, with equality only if $ab \geq 0$ |
| $$|x| = - 3 \Rightarrow x = \pm 3$$ | $|x| \geq 0$, so the equation has no solutions |
| $|x - 3| = 2x \Rightarrow x - 3 = \pm 2x$, with no other condition | first the condition $2x \geq 0$, then checking the solutions |
| $$|x| < a \Rightarrow x < a$$ | $|x| < a \Leftrightarrow - a < x < a$ ($a > 0$) |
| $$\left\lbrack - \text{3,4} \right\rbrack = - 3$$ | $\left\lbrack - \text{3,4} \right\rbrack = - 4$ and $\left\{ - \text{3,4} \right\} = \text{0,6}$ |
| $$\lbrack 2x\rbrack = 2\lbrack x\rbrack$$ | false in general: $\left\lbrack 2 \cdot \text{0,5} \right\rbrack = 1$, but $2\left\lbrack \text{0,5} \right\rbrack = 0$ |
| $$\lbrack x + y\rbrack = \lbrack x\rbrack + \lbrack y\rbrack$$ | true only if $\left\{ x \right\} + \left\{ y \right\} < 1$ |
6. General algorithm for equations and inequalities
- Identify the type of equation/inequality (the tables in 3.4, 3.5 and 4.4).
- Write the conditions: right-hand side $\geq 0$ (for $\left| E(x) \right| = F(x)$), $k \in {\mathbb{Z}}$ (for the integer part), $f \in \lbrack 0,\ 1)$ (for the fractional part).
- Solve by cases or by intervals.
- Intersect the solutions of each case with the interval or condition of that case.
- Combine the solutions and write the set S (as a finite set or as an interval/union of intervals).
- Check by substituting at least one solution.