Let a and b be real numbers. If a – b is negative we say that a is less than b and write a < b. If a – b is positive then a is greater than b, i.e., a > b.

1. For any two real numbers a and b, we have a > b or a = b or a < b.

2. If a > b and b > c, then a > c. If a > b then (a + c) > (b + c) and (a - c) > (b - c), however, ac > bc and (a/c) > (b/c) (not sure) (is true only when c is positive)

3. If a > b, then a + m > b + m, for any real number m.

4. If a ≠ 0, b ≠ 0 and a > b, then \(\frac{1}{a} < \frac{1}{b}\).

5. If a > b, then am > bm for m > 0 and am < bm for m < 0, that is, when we multiply both sides of inequality by a negative quantity, the sign of inequality is reversed.

6. If a > X, b > Y, c > Z then (1) a + b + c + .... > X + Y + Z + .... (2) abc .... > XYZ .... (Provided none is negative)

7. If x > 0 and a > b > 0, then \(a^x > b^x\)

8. If a > 1 and x > y > 0, then \(a^x > a^y\)

9. If 0 < a < 1 and x > y > 0, then \(a^x < a^y\)

10. Do not cancel anything from both sides of inequality unless you are sure that the canceled quantity is positive, so ax > ay does not necessarily mean x > y, etc.

11. The concept of number line is very useful in checking inequalities. The common values to check are x = 0, 1, -1, >1 (preferred value = 2), between 0 and 1 (preferred value = 1/2), between - 1 and 0 (preferred value = -1/2), and less than -1 (preferred value = -2). So in short, there are 7 points: -2, -1, -1/2, 0, 1/2, 1, 2.

12. |x| is defined as the non-negative value of x and hence is never negative. On the GRE, \(\sqrt{x^2} = |x|\), that means, the square root of any quantity is defined to be non-negative, so \(\sqrt{36} = 6\) and not − 6 on the GRE. BUT if \(x^2=36\) ⇒ \(x=6\) or \(-6\) both. So \(\sqrt{x^2}= x\) or \(-x\) both are possible. If, x is negative, then \(\sqrt{x^2}=-x\) as it has to be +ve eventually. In this case x is negative and -x is positive.

13. |5| = 5, |-5| = 5, so |x| = x, if x is positive or 0 and |x| = -x if x is negative.

14. If |x| > x, then x is negative.

15. If |x| = a, then x = a or x = -a.

16. If |x| > a, then x > a or x < -a.

17. If |x| < a, then x < a or x > -a.

18. If |x - a| > b, then either x - a > b or x - a < -b

19. If |x - a| < b, then either x - a < b or x - a > -b.

20. If |x| = x, then x is either positive or 0.

21. |a + b| ≤ |a| + |b|, |a – b| ≥ ||a| – |b||, |ab| = |a| |b|, \(|\frac{a}{b}|=|\frac{a}{b}|\), \(b \neq 0\), \(|a^2|=a^2\)

22. If (x - a) (x - b) < 0, then x lies between a and b. OR a < x < b.

23. If (x - a) (x - b) > 0, then x lies outside a and b. OR x < a, x > b.

24. If \(x^2> x\), then either x > 1 or x is negative (x < 0)

25. If \(x^2< x\), then x lies between 0 and 1. (0 < x < 1)

26. If \(x^2= x\), then x = 0 or x = 1.

27. If \(x^3> x\), then either x > 1 or x is between -1 and 0(either x > 1 or -1 < x < 0).

28. If \(x^3< x\), then either x lies between 0 and 1 or x is less than -1. (either 0 < x < 1 or x < -1)

29. If \(x^3= x\), then x = 0 or x = 1 or x = -1.

30. If \(x^3= x\), then x = 0 or x = 1 or x = -1.

31. If x > y, it is not necessary that \(x^2> y^2\) or \(\sqrt{x} > \sqrt{y}\) etc. So even powers can’t be predicted.

32. If x > y, it is necessarily true that \(x^3 > y^3\) or \(\sqrt[3]{x} > \sqrt[3]{y}\) etc. So odd powers and roots dont change sign.

33. ab > 0 means \(\frac{a}{b} > 0\) and vice versa. The two are of the same sign.

34. ab < 0 means \(\frac{a}{b} < 0\) and vice versa. The two are of the opposite sign.

35. If x is positive, \(x + \frac{1}{x} ≥ 2\) .

36. If X is positive, then

(1) \(\frac{(a + X) }{ (b + X)} > \frac{a}{b}\) if a < b

(2) \(\frac{(a + X) }{ (b + X)} < \frac{a}{b}\) if a > b

37. If X is negative, then

(1) \(\frac{(a + X) }{ (b + X)} > \frac{a}{b}\) if a > b

(2) \(\frac{(a + X) }{ (b + X)} < \frac{a}{b}\) if a < b

38. \(\frac{(a + c + e + ....) }{ (b + d + f + ....)}\) is less than the greatest and greater than the least of the fractions \(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}, .....
\)

39.

The solution set of an inequality is a set of numbers, each element of which, when substituted for the variable, results in a true inequality.




40.

Properties of Inequality

Addition properties of inequality:

- If $\(A<B\)$, then $\(A+C<B+C\)$
- If $\(A>B\)$, then $\(A+C>B+C\)$

Subtraction properties of inequality:

- If $\(A<B\)$, then $\(A-\mathrm{C}<B-\mathrm{C}\)$
- If $\(A>B\)$, then $\(A-\mathrm{C}>B-\mathrm{C}\)$

Multiplication and division by a positive number:

- If $\(A<B\)$ and $\(\mathrm{C}>0\)$, then $\(A C<B C\)$
- If $\(A>B\)$ and $\(C>0\)$, then $\(A C>B C\)$
- If $\(A<B\)$ and $\(C>0\)$, then $\(A / C<B / C\)$
- If $\(A>B\)$ and $\(C>0\)$, then $\(A / C>B / C\)$

These properties also apply to $\leq$ and $\geq$.

- If $\(A \leq B\)$, then $\(A+C \leq B+\mathrm{C}\)$
- If $\(A \geq B\)$, then $\(A+C \geq B+\mathrm{C}\)$
- If $\(A \leq B\)$, then $\(A-C \leq B-\mathrm{C}\)$
- If $\(A \geq B\)$, then $\(A-C \geq B-\mathrm{C}\)$

Multiplication and division by a negative number:

- If $\(A<B\)$ and $\(C<0\)$, then $\(A C>B C\)$
- If $\(A>B\)$ and $\(C<0\)$, then $\(A C<B C\)$
- If $\(A<B\)$ and $\(C<0\)$, then $\(A / C>B / C\)$
- If $\(A>B\)$ and $\(C<0\)$, then $\(A / C<B / C\)$

These properties are also applicable to $\(\leq\)$ and $\(\geq\)$.


41.

The Absolute value of a number $x$ is written $|x|$ and is defined as

$$
\(|x|=x \text { if } x \geq 0 \quad \text { or } \quad|x|=-x \text { if } x<0\)
$$


That is, $\(|4|=4\)$ since 4 is positive, but $\(|-2|=2\)$ since -2 is negative.

We can also think of $\(|x|\)$ geometrically as the distance of $\(x\)$ from 0 on the number line.



\(\text { More generally, }|x-a| \text { can be thought of as the distance of } x \text { from } a \text { on the numberline. }\)





Note that $\(|a-x|=|x-a|\)$.

The absolute value function is written as $\(y=|x|\)$.

We define this function as

$$
\(y= \begin{cases}+x & \text { if } x \geq 0 \\ -x & \text { if } x<0\end{cases}\)
$$


From this definition we can graph the function by taking each part separately. The graph of $y=|x|$ is given below.





\(\text { The graph of } y=|x| \text {. }\)


Mode Function





Example

For what values of $x$ is $\(|x-4|=|2 x-1|\)$.

Solution

We know that the values $\(x=\frac{1}{2}\)$ and $\(x=4\)$ are important $x$ values here, so we will use them to divide the $\(x\)$ axis into three sections and will consider them in turn.

Case 1.

For $\(x<\frac{1}{2},|x-4|=-(x-4)=|2 x-1|=-(2 x-1)\)$, so $\(-x+4=-2 x+1\)$.
Therefore, $\(x=-3\)$.

Case 2.

For $\(\frac{1}{2} \leq x<4,|x-4|=-(x-4)=|2 x-1|=2 x-1\)$, so $\(-x+4=2 x-1\)$.
Therefore, $\(x=\frac{5}{3}\)$.

Case 3.

For $\(x \geq 4,|x-4|=x-4=|2 x-1|=2 x-1\)$, so $\(x-4=2 x-1\)$.
Therefore, $\(x=-3\)$, but this does not satisfy the assumption $x \geq 4$ so this case does not give us a solution.
The solutions are $\(x=-3\)$ and $\(x=\frac{5}{3}\)$.




Watch the following video to learn the Basics of Inequalities



Watch the following video to learn How to Solve Inequality Problems







Source used:

- http://www.alamo.edu
- Mathematics Learning Centre The University of Sidney
https://www.sydney.edu.au/students/lear ... maths.html