Order and translation
I- Comparing two rational numbers
Définition :
▶ Comparing two rational numbers means : determining which one of the two numbers is the smaller and which is the bigger
Rule :
➤ To compare two rational numbers $\mathbf{a}$ and $\mathbf{b}$ : we determine the sign of their difference $\mathbf{a – b}$ (or b-a)
➤ if $a-\mathbf{b}>\mathbf{0}($ or $\mathbf{b}-\mathbf{a}<\mathbf{0})$ Then $a>b($ or $b<a)$
➤ if $a-\mathbf{b}<0 \quad($ or $\mathbf{b}-\mathbf{a}>0)$ Then $a<b \quad($ or $b>a)$
Rule :
➤ We know the order of two rational numbers $\mathbf{a}$ and $\mathbf{b}$ by determining the sign of their difference
➤ if $a$ – $\mathbf{b}$ is positive Then $a>b$
➤ if $a$ – $\mathbf{b}$ is negative Then $a<b$
Exemple :
– We have : $23-35=-12<0$
– So : $23<35$
We have : $\frac{5}{6}-\frac{4}{5}=\frac{5 \times 5}{6 \times 5}-\frac{4 \times 6}{6 \times 5}=\frac{25}{30}-\frac{24}{30}=\frac{1}{30}>0$
So : $\frac{5}{6}>\frac{4}{5}$
Rule :
– if $a-\mathbf{b}=\mathbf{0}($ or $\mathbf{b}-\mathbf{a}=\mathbf{0})$ Then $a=b$
Exemple :
– Let’s compare the two numbers : $2 x+\frac{1}{2}$ and $2 x+\frac{1}{3}$
– We have : $\begin{aligned}\left(2 x+\frac{1}{2}\right)-\left(2 x+\frac{1}{3}\right)= & 2 x+\frac{1}{2}-2 x-\frac{1}{3} \\ & =\frac{1}{2}-\frac{1}{3}=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\end{aligned}$
– Since: $\frac{1}{6}>0$ then : $\left(2 x+\frac{1}{2}\left(2 x+\frac{1}{3}\right)>0\right.$
– So : $\left(2 x+\frac{1}{2}>\left(2 x+\frac{1}{3}\right)\right.$
1. Vocabulary :
The symbol <
– a, b are rational numbers :
– We have : if a < b meansainferiorb
– We read : $\mathbf{a}$ is strictly inferior to $\mathbf{b}$
The symbol $\leq$
– $\mathbf{a , b}$ are rational numbers :
– We have : if $a \leq$ bmeansainferiorb
– We read : $\mathbf{a}$ is inferior or equal to $\mathbf{b}$
The symbol >
– The symbol > means strictly superior
– $\mathbf{a , b}$ are rational numbers :
– We have : if $a>$ bmeansasuperiorb
– We read : $\mathbf{a}$ is strictly superior to $\mathbf{b}$
The symbol $\geq$
– The symbol $\geq$ means superior or equal
– $\mathbf{a , b}$ are rational numbers :
– We have : if $a \geq$ bmeansasuperiorb
– We read : $\mathbf{a}$ is superior or equal to $\mathbf{b}$
2. Inequality :
Définition :
▶ An inequality is a mathematical sentence that is used to compare quantities
▶ Inequalities are : two expressions separated by one of the four inequality signs $\langle, \leqslant \geqslant$, or $\geqslant$.
Exemple :
We consider the following table :
Définition :
▶ $\mathbf{a , b}$ are rational numbers :
▶ Every expression in the form of $\mathbf{a}<\mathbf{b}$ or $\mathbf{a}>\mathbf{b}$ or $\mathbf{a}<\mathbf{b}$ or $\mathbf{a}>\mathbf{b}$ is called an inequality
▶ The numbers a and b are called the inequality sides
II-Order and operations
1. Order and addition :
Propriety 1 :
We have : if $a \leq b$ thena $+k \leq b+k$
if $a \geq b$ Thena $+k \geq b+k$
Exemple :
– $x$ and $y$ are rational numbers such as : $x \leq 7$ and $y \leq 5$
So : $\quad x+y \leq 7+5$
Then : $x+y \leq 12$
Propriety 2 :
We have : if $a \leq b \wedge c \leq$ dthen $a+c \leq b+d$ if $a \geq b \wedge c \geq d$ Thena $+c \geq b+d$
Exemple :
– $x$ and $y$ are rational numbers such as : $x+1 \leq \frac{7}{3}$ and $y-1 \leq-2$
– Demonstrate that : $x+y \leq \frac{1}{3}$
– We have : $\quad x+1 \leq \frac{7}{3}$ and $y-1 \leq-2$
So : $x+1+y-1 \leq \frac{7}{3}-2$
Then : $x+y \leq \frac{7}{3}-\frac{6}{3}$
$x+y \leq \frac{1}{3}$
2. Order and subtract :
Propriety 2 :
$\mathbf{a , b}$ and $\mathbf{K}$ are rational numbers :
$\begin{aligned} & \text { We have : } \text { if } a \leq b \text { then } a-k \leq b-k \\ & \qquad \text { if } a \geq b \text { Then } a-k \geq b-k\end{aligned}$
Exemple :
– x is a rational number such as: $\mathrm{x} \leq 7$
So : $\quad x-4 \leq 7-4$
Then : $x-4 \leq 3$
3. Order and multiplication :
Propriety 1 :
$\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are rational numbers :
We have : if $a \leq b \wedge c>0$ then $a \times c \leq b \times c$
if $a \leq b \wedge c<0$ Then $a \times c \geq b \times c$
Exemple :
– $\mathbf{a}$ and $\mathbf{b}$ are rational numbers :
If $a \geq \frac{4}{3}$
So : $5 a \geq \frac{20}{3}$
If $b \leq 3$
So : $-4 b \geq-12$
Propriety 2 :
$\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ and $\mathbf{d}$ are positives rational numbers :
We have : if $a \leq b \wedge c \leq$ dthena $\times c \leq b \times d$
4. Order and the opposite
Propriety :
– a, b rational numbers :
We have : if $a \leq b$ then $-a \geq-b$
5. Order and the inverse :
Propriety :
$\mathbf{a , b}$ rational numbers non null and have the same sign :
We have : if $a \leq b$ then $\frac{1}{a} \geq \frac{1}{b}$
Exemple :
– $\mathbf{a}$ and $\mathbf{b}$ are non-null rational numbers :
– We have : $a=5$ and $b=3$
– Since : $5 \geq 3 \quad$ So : $\quad \frac{1}{3} \geq \frac{1}{5}$
– We have: $a=(-1)$ and $b=(-2)$
– Since : (-1) $\geq(-2)$
– So $\frac{1}{-2} \geq \frac{1}{-1}$
– Then : $\frac{1}{-2} \geq-1$
6. Order and the square :
Propriety :
$\mathbf{a}, \mathbf{b}$ are rational numbers positive :
We have : if $^2 \leq$ bthen $^2 \leq b^2$
Exemple :
– $\mathbf{a}$ and $\mathbf{b}$ are rational numbers
– We have : $a=5$ and $b=3$
– So : $a^2=5^2=25$ and $b^2=3^2=9$
– Since $a \geq b$ then $a^2 \geq b^2$
Propriety :
b, $\mathbf{b}$ are rational numbers negative :
We have: if $a \leq b$ then $a^2 \geq b^2$
Exemple :
– $\mathbf{a}$ and $\mathbf{b}$ are rational numbers
– We have $\mathrm{a}=(-2)$ and $\mathrm{b}=(-3)$
– So : $\quad a^2=(-2)^2=4 \quad$ and $\quad b^2=(-3)^2=9$
– Since $a \geq b$ then $a^2 \leq b^2$
III- Framing of a rational number
Définition :
► $\mathbf{a , b}$ and $\mathbf{x}$ are rational numbers :
$$
\text { if } a \leq x \wedge x \leq b \text { then } a \leq x \leq b
$$
► The writing $a \leq x \leq b$ is called a framing of the number x
► All the expressions : $a \leq x \leq b$ and $a<x<b$ and $a \geq x \geq b$ and $a>x>b$ are called a framing of the number x
Remark :
– $a \leq x \leq b$ is read $\mathbf{x}$ is between $\mathbf{a}$ and $\mathbf{b}$
– $a<x<b$ is read $\underline{\mathbf{x}}$ is strictly between $\mathbf{a}$ and $\mathbf{b}$
– $a \leq x \leq b$ means that : $a \leq x$ and $x \leq b$
1. Framing and addition :
Propriety :
a, b, c, d, x, y are rational numbers :
We have :
$$
\text { if: }\left\{\begin{array}{l}
a \leq x \leq b \\
c \leq y \leq d
\end{array} \quad \text { Then } a+c \leq x+y \leq b+d\right.
$$
Exemple :
$\mathbf{a}$ and $\mathbf{b}$ are non-null rational numbers :
– We have : $2 \leq a \leq 3$ and $\frac{1}{2} \leq b \leq \frac{3}{2}$
– Let’s frame : $\mathrm{a}+\mathrm{b}$
– We have : $2+\frac{1}{2} \leq a+b \leq 3+\frac{3}{2}$
– So : $\quad \frac{4}{2}+\frac{1}{2} \leq a+b \leq \frac{6}{2}+\frac{3}{2}$
– Then : $\quad \frac{5}{2} \leq a+b \leq \frac{9}{2}$
2. Framing and the opposite :
Propriety :
a, b, x are rational numbers :
We have :
if $\quad a \leq x \leq b \quad$ Then $\quad-b \leq-x \leq-a$
Exemple :
$\mathbf{a}$ and $\mathbf{b}$ are non-null rational numbers :
– We have : $1 \leq a \leq 2$ and $-4 \leq b \leq-3$
– Let’s frame : (-a) and (-b)
– We have : $1 \leq a \leq 2$
– So $2 \times(-1) \leq a \times(-1) \leq 1 \times(-1)$
– Then : $\quad-2 \leq-a \leq-1$
– We have : $-4 \leq b \leq-3$
– So $(-3) \times(-1) \leq b \times(-1) \leq(-4) \times(-1)$
– Then : $\quad 3 \leq(-b) \leq 4$
3. Framing and subtract :
Remark :
▸ We know : $a-b=a+(-b)$ so to frame the number $a-b$ ! should frame first (-b) then use the propriety of framing and addition on $\mathrm{a}+(-\mathrm{b})$
Propriety :
a, b, x are rational numbers :
We have :
if $\left\{\begin{array}{l}a \leq x \leq b \\ c \leq y \leq d\end{array}\right.$ Then $a-d \leq x-y \leq b-c$
Exemple :
$\mathbf{a}$ and $\mathbf{b}$ are non-null rational numbers :
– We have : $4 \leq a \leq 7$ and $-3 \leq b \leq-2$
– Let’s frame : a-b
– We have : $-3 \leq b \leq-2$
– So $(-2) \times(-1) \leq b \times(-1) \leq(-3) \times(-1)$
– Then : $\quad 2 \leq-b \leq 3$
– So : $\quad 2+4 \leq a-b \leq 3+7$
– Then : $\quad 6 \leq a-b \leq 10$
4. Framing and multiplication :
Propriety :
$\mathbf{a , b , x}$ and $\mathbf{k}$ are rational numbers such as : $a \leq x \leq b$
if $\boldsymbol{k}<\mathbf{0} \quad$ Then $\boldsymbol{k} \times \boldsymbol{a} \leq \boldsymbol{K} \times \boldsymbol{x} \leq \boldsymbol{k} \times \boldsymbol{b}$
if $\quad \boldsymbol{k}>\mathbf{0} \quad$ Then $\boldsymbol{k} \times \boldsymbol{b} \leq \boldsymbol{K} \times \boldsymbol{x} \leq \boldsymbol{k} \times \boldsymbol{a}$
Exemple :
$\mathbf{a}$ is a rational number such as: $4 \leq a \leq 7$
– Let’s frame : 2a and -3a
– We have : $\quad 4 \leq a \leq 7$
– So $\quad 2 \times 4 \leq 2 \times a \leq 7 \times 2$
– Then : $\quad 8<2 a<14$
– So : $\quad(-3) \times 7 \leq(-3) \times a \leq 4 \times(-3)$
– Then : $\quad-21 \leq-3 a \leq-12$
Exemple 2 :
$\mathbf{a}$ and $\mathbf{b}$ are rational numbers such as : $4 \leq a \leq 7$ and $-3 \leq b \leq-2$
– Let’s frame : $\mathrm{a} \times \mathrm{b}$
– $1^{\text {st }}$ Let’s frame : $-b$ because $b<0$
– We have : $(-2) \times(-1) \leq b \times(-1) \leq(-3) \times(-1)$
– Then : $\quad 2 \leq-b \leq 3$
– We have : $2 \times 4 \leq a \times(-b) \leq 3 \times 7$
– So $\quad 8 \leq-a b \leq 21$
– Then : $21 \times(-1) \leq(-a b) \times(-1) \leq 8 \times(-1)$
– So : $\quad-21 \leq a \times b \leq-8$
5. Framing and the inverse :
Propriety :
a, b, x are rational numbers :
We have :
if $\quad \boldsymbol{a} \leq \boldsymbol{x} \leq \boldsymbol{b} \quad$ Then $\quad \frac{1}{b} \leq \frac{1}{x} \leq \frac{1}{a}$
Exemple :
$\mathbf{a}$ is a non-null rational number such as : $4 \leq a \leq 7$
– We have : $4 \leq a \leq 7$
– So $\quad \frac{1}{7} \leq \frac{1}{a} \leq \frac{1}{4}$
b is a non-null rational number such as : $-3 \leq b \leq-2$
– We have : $-3 \leq b \leq-2$
– So $\quad \frac{1}{-2} \leq \frac{1}{b} \leq \frac{1}{-3}$
6. Framing and the square :
Propriety :
$\mathbf{a , b , x}$ are rational numbers such as : $a \leq x \leq b$
We have : $\quad$ if $\mathrm{x}>0$ then $a^2 \leq x^2 \leq b^2$
if $x<0$ then $b^2 \leq x^2 \leq a^2$
Exemple :
$\mathbf{a}$ and $\mathbf{b}$ are non-null rational numbers such a : $1 \leq \mathbf{a} \leq 2$ and $-3 \leq b \leq-2$
– We have : $1^2 \leq a^2 \leq 2^2$
– So $1 \leq a^2 \leq 4$
– We have : $(-2)^2 \leq b^2 \leq(-3)^2$
– So
$4 \leq b^2 \leq 9$
IV- Inequations
Définition :
► $\mathbf{a , b}$ and $\mathbf{x}$ are rational numbers:
► The writing $a x+b \leq 0$ is called a 1 st degree inequation with one variable
Exemple :
– x is a rational number.
– We have : $5 x \leq 15 ; ; x+\frac{3}{2}>0 ; ; 25 x-1 \leq 0 ; ; \frac{7}{2} x+\frac{2}{5}<3 x+1$ are 1 st degree inequations with one variable,
Exemple 2 :
– x is a rational number such as :
– Solve the following inequation : $2 x-1 \geq 5$
– We have : $\quad 2 x-1 \geq 5$
– So : $\quad 2 x \geq 5+1$
– Then : $\quad 2 x \geq 6$
– So : $\quad \frac{1}{2} \times 2 x \geq 6 \times \frac{1}{2}$
– Finally : $\quad x \geq 3$
– So : All the rational numbers superior or equal to 3 are solution to this inequation
Remark :
▸Many problems can be translated into inequations. To solve problems using algebra, we follow these steps :
– Step 1: Decide the unknown quantity and allocate it a variable such as x.
– Step 2 : Translate the problem into an inequation
– Step 3 : Solve the inequation by isolating the variable.
– Step 4 : Check that the solution satisfies the original problem and write the answer in sentence form, describing how the solution relates to the original problem.
Exemple :
– A Game Park offers two membership formulas :
– Formula A : the annual pass costs 550DH and the price of an entry is 200 DH .
– Formula B : the annual pass costs 800 DH and the price of an entry is 150 DH .
– The number of admissions is x .
– From how many entries in one year is formula B the most interesting?
Solution :
Step 1: Define the Unknown Variable :
– Let $\mathbf{x}$ represents the number of admissions (entries) in one year.
– So the Annual Cost for Formula A is: 550+200x
– the Annual Cost for Formula B is: $\mathbf{8 0 0 + 1 5 0 x}$
– Step 2 : Translate to an inequation :
– We want to find the minimum number of entries $\mathbf{x}$ where Formula B Becomes More Interesting (Cheaper) Than Formula A.
$800+150 x<550+200 x$
– Step 3 : Solve the inequation
– We have : $\quad \mathbf{8 0 0 + 1 5 0 x < 5 5 0 + 2 0 0 x}$
– So : $\quad 800-550<200 x-150 x$
– Then : 250<50x
– Finaly : $\quad 5<x$
– Step 4 : Conclusion
$>$ if $x \leq 5$ : Formula $A$ is cheaper or equal.
If $x>5$ : Formula B becomes cheaper.
V- VOCABULARY
