What are the different types of numbers in mathematics?

The main types are: natural numbers N = {1, 2, 3, ...}, whole numbers W = {0, 1, 2, ...}, integers Z = {..., -1, 0, 1, ...}, rational numbers Q = {p/q : q ≠ 0}, irrational numbers (like π, √2), real numbers R (all rationals plus irrationals), and complex numbers C = {x + iy : x, y ∈ R}. They form a nested hierarchy: N ⊂ W ⊂ Z ⊂ Q ⊂ R ⊂ C.

What is the difference between rational and irrational numbers?

A rational number can be written as p/q with p, q integers and q ≠ 0; its decimal expansion is either terminating or non-terminating repeating. An irrational number cannot be written as such a ratio; its decimal is non-terminating and non-repeating. Examples of irrational: π, √2, e, log₁₀ 3.

What is the wavy curve method?

The wavy curve method is a technique for solving rational polynomial inequalities like (x - a₁)(x - a₂).../(x - b₁)(x - b₂)... ≥ 0. Mark critical points on the number line, check the sign at one convenient point, then draw a wavy curve that crosses at odd-multiplicity roots and touches (stays on the same side) at even-multiplicity roots. The intervals matching the desired sign form the solution.

How do you solve |x - a| < b?

For b > 0, |x - a| b, the answer is x a + b.

What is the fundamental logarithmic identity?

The fundamental identity is a^(log_a b) = b, valid for a > 0, a ≠ 1, b > 0. It says: raising a to the power log_a b gives back b. This identity follows directly from the definition log_a b = x iff a^x = b.

What are the basic properties of logarithms used in JEE?

The core properties are: log_a(mn) = log_a m + log_a n, log_a(m/n) = log_a m - log_a n, log_a m^n = n log_a m, log_a 1 = 0, log_a a = 1, change of base log_a b = log_c b / log_c a, and log_b a = 1/log_a b. Base sign matters: log_a x is increasing for a > 1 and decreasing for 0 < a < 1.

Why does log_a x require a > 0 and a ≠ 1?

For a to serve as a base, a^x must be well-defined and take all positive real values. If a ≤ 0, a^x is not defined for all real x (e.g. a^(1/2) fails for a 0 for all real y.

What are the four types of finite intervals?

Given a < b: open (a, b) = {x : a < x < b} (both endpoints excluded); closed [a, b] = {x : a ≤ x ≤ b} (both included); open-closed (a, b] = {x : a < x ≤ b} (right endpoint included); closed-open [a, b) = {x : a ≤ x < b} (left endpoint included). Infinite intervals extend to ±∞ and always use round brackets at ∞.

How do you handle inequalities when multiplying by a negative number?

Multiplying (or dividing) both sides of an inequality by a negative number reverses the direction of the inequality. For example, -2x > 4 becomes x < -2 after dividing by -2. If the sign of what you're multiplying by is unknown (like a variable expression), split into cases based on that sign, or move everything to one side and use the wavy curve method.

Number System And Logarithms

Q: How do you solve |x - a| < b?

For b > 0, |x - a| b, the answer is x a + b.

Q: What are the four types of finite intervals?

Given a < b: open (a, b) = {x : a < x < b} (both endpoints excluded); closed [a, b] = {x : a ≤ x ≤ b} (both included); open-closed (a, b] = {x : a < x ≤ b} (right endpoint included); closed-open [a, b) = {x : a ≤ x < b} (left endpoint included). Infinite intervals extend to ±∞ and always use round brackets at ∞.

Q: How do you handle inequalities when multiplying by a negative number?

Multiplying (or dividing) both sides of an inequality by a negative number reverses the direction of the inequality. For example, -2x > 4 becomes x < -2 after dividing by -2. If the sign of what you're multiplying by is unknown (like a variable expression), split into cases based on that sign, or move everything to one side and use the wavy curve method.

MathsSets, Relations And Number SystemFor JEE aspirants

The number system classifies numbers into natural numbers $N$ , whole numbers $W$ , integers $Z$ , rational $Q$ , irrational $Q^{c}$ , real $R$ , and complex $C$ . From this foundation come intervals, inequalities (including the powerful wavy curve method for rational polynomial inequalities), absolute value, and logarithms. This page is a full JEE-level walkthrough of all these topics with 14 solved examples and clear property lists.

Key Formulas - Quick Reference

Hierarchy: $N \subset W \subset Z \subset Q \subset R \subset C$
AM-GM style: $a + \frac{1}{a} \geq 2$ for $a > 0$ (equality at $a = 1$ ); $a + \frac{1}{a} \leq - 2$ for $a < 0$ (equality at $a = - 1$ )
Absolute value: $∣ x ∣ = x^{2} = max {x, - x}$
$∣ x ∣ < a ⟺ - a < x < a$ (for $a > 0$ ); $∣ x ∣ > a ⟺ x > a$ or $x < - a$
Fundamental log identity: $a^{l o g_{a} b} = b$ (for $a > 0, a \neq = 1, b > 0$ )
$lo g_{a} (mn) = lo g_{a} m + lo g_{a} n$ ; $lo g_{a} (\frac{m}{n}) = lo g_{a} m - lo g_{a} n$ ; $lo g_{a} m^{n} = n lo g_{a} m$
Change of base: $lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}$ ; $lo g_{b} a = \frac{1}{lo g _{a} b}$
Base rules: for $b > 1$ , $lo g_{b} x$ is increasing; for $0 < b < 1$ , decreasing

1. Number System

Natural Numbers ( $N$ )

The counting numbers $1, 2, 3, 4, \dots$ form the set of natural numbers: $N = {1, 2, 3, 4, 5, \dots}$ .

Whole Numbers ( $W$ )

Natural numbers together with $0$ : $W = {0, 1, 2, 3, 4, \dots}$ .

Integers ( $Z$ or $I$ )

All whole numbers plus their negatives: $Z = {\dots, - 3, - 2, - 1, 0, 1, 2, 3, \dots}$ .

Positive integers: $Z^{+} = {1, 2, 3, \dots}$ (same as $N$ )
Negative integers: $Z^{-} = {\dots, - 3, - 2, - 1}$
Non-negative integers: ${0, 1, 2, 3, \dots}$ (same as $W$ )
Non-positive integers: ${\dots, - 3, - 2, - 1, 0}$

Rational Numbers ( $Q$ )

Numbers of the form $\frac{p}{q}$ where $p, q \in Z$ and $q \neq = 0$ (with $g cd (p, q) = 1$ in reduced form):

$Q = {\frac{p}{q} : p, q \in Z, q \neq = 0} .$

Every integer is rational (write $n = n /1$ ). Rationals are exactly the real numbers whose decimal expansion is either terminating (e.g. $3/4 = 0.75$ ) or non-terminating repeating (e.g. $23/11 = 2. \overline{09}$ ).

Irrational Numbers ( $Q^{c}$ )

Real numbers that are not rational. Their decimal expansions are non-terminating and non-repeating. Examples: $π$ , $2$ , $35$ , $lo g_{10} 3$ .

Real Numbers ( $R$ )

The union of rationals and irrationals: $R = Q \cup Q^{c}$ . Every point on the number line is a real number.

Complex Numbers ( $C$ )

Numbers of the form $z = x + i y$ where $x, y \in R$ and $i = - 1$ . Here $x$ is the real part and $y$ the imaginary part. Every real number is complex (with $y = 0$ ), so $R \subset C$ .

Figure: Real number line showing the hierarchy N ⊂ W ⊂ Z ⊂ Q ⊂ R.

2. Intervals

A subset of the real line described by inequalities is called an interval. Intervals are essential when solving inequalities and describing domains. If $a, b \in R$ with $a < b$ :

Finite Intervals

Type	Notation	Set
Open	$(a, b)$	${x : a < x < b}$
Closed	$[a, b]$	${x : a \leq x \leq b}$
Open-closed	$(a, b]$	${x : a < x \leq b}$
Closed-open	$[a, b)$	${x : a \leq x < b}$

Infinite Intervals

Notation	Set
$(a, \infty)$	${x : x > a}$
$[a, \infty)$	${x : x \geq a}$
$(- \infty, b)$	${x : x < b}$
$(- \infty, b]$	${x : x \leq b}$
$(- \infty, \infty)$	$R$

Figure: The four types of finite intervals plus infinite intervals on the real line.

3. Inequalities

If $p, q, r$ are real numbers, the basic rules of inequalities are:

$p \geq q \Rightarrow$ either $p > q$ or $p = q$
$p < q \Rightarrow p + r < q + r$ (adding same quantity preserves inequality)
$p < q$ and $q < r \Rightarrow p < r$ (transitivity)
For $p > q$ : $k p > k q$ if $k > 0$ , and $k p < k q$ if $k < 0$ (multiplying by negative flips the sign)
For $0 < p < q$ : $p^{r} < q^{r}$ if $r > 0$ , and $p^{r} > q^{r}$ if $r < 0$
$a + \frac{1}{a} \geq 2$ for all $a \in R^{+}$ , equality at $a = 1$
$a + \frac{1}{a} \leq - 2$ for all $a \in R^{-}$ , equality at $a = - 1$

Inequalities Involving Exponentials

If $k > 0$ , then $k^{x} > 0$ for all real $x$
If $k > 1$ , then $k^{x} > 1$ when $x > 0$
If $0 < k < 1$ , then $k^{x} < 1$ when $x > 0$ , and $k^{x} > 1$ when $x < 0$

4. Wavy Curve Method

The wavy curve (sign chart) method solves inequalities of the form:

$f (x) = \frac{( x - a _{1} ) ^{n_{1}} ( x - a _{2} ) ^{n_{2}} \dots ( x - a _{k} ) ^{n_{k}}}{( x - b _{1} ) ^{m_{1}} ( x - b _{2} ) ^{m_{2}} \dots ( x - b _{p} ) ^{m_{p}}} \geq 0 (or > 0, < 0, \leq 0),$

where the $n_{i}, m_{j}$ are natural numbers and all $a_{i} \neq = b_{j}$ .

The Method - Step by Step

Arrange all values at which the numerator or denominator becomes zero (i.e. $a_{1}, \dots, a_{k}, b_{1}, \dots, b_{p}$ ) in increasing order. Plot them on the real line.
Mark points where the numerator is zero with solid (filled) circles.
Mark points where the denominator is zero with hollow (empty) circles (these are excluded).
Check the sign of $f (x)$ for one value larger than the right-most critical point (usually easy).
Draw a wavy curve from right to left through all critical points. At a point whose exponent is odd, the curve crosses the number line (sign changes). At a point whose exponent is even, the curve touches and stays on the same side (sign does not change).
Read off intervals: $f (x) > 0$ where the curve is above the line; $f (x) < 0$ where below. Take the union of intervals matching the inequality.

Figure: Wavy curve method - sign alternates at odd-multiplicity roots, stays same at even-multiplicity roots.

Worked Example

Example

Solve

\frac{( x + 1 ) ^{2} ( x - 2 ) ^{3} ( x - 4 ) ^{4}}{( x - 3 ) ^{2} ( x - 7 ) ^{5}} \geq 0

Solution:

Numerator zeros (marked with solid dots, included since we allow equality): $x = - 1$ (multiplicity $2$ , even), $x = 2$ (multiplicity $3$ , odd), $x = 4$ (multiplicity $4$ , even).

Denominator zeros (marked with hollow dots, excluded): $x = 3$ (multiplicity $2$ , even), $x = 7$ (multiplicity $5$ , odd).

Ordered on the number line: $- 1, 2, 3, 4, 7$ .

Check sign at $x = 8$ : every factor is positive, so $f (8) > 0$ . Draw the wavy curve from top-right leftward:

Figure: Sign chart for the example - curve touches at -1, 3, 4 (even) and crosses at 2, 7 (odd).

At $x = 7$ (odd): curve crosses. Sign flips to negative on $(4, 7)$ .
At $x = 4$ (even): curve touches, stays negative on $(3, 4)$ .
At $x = 3$ (even): curve touches, stays negative on $(2, 3)$ .
At $x = 2$ (odd): curve crosses. Sign flips to positive on $(- 1, 2)$ .
At $x = - 1$ (even): curve touches, stays positive on $(- \infty, - 1)$ .

Where is $f (x) \geq 0$ ? Above the axis: $(- \infty, - 1) \cup (- 1, 2) \cup (7, \infty)$ . Include numerator zeros (equality allowed): add $x = - 1$ , $x = 2$ , $x = 4$ . Exclude denominator zeros: $x = 3, 7$ .

$x \in (- \infty, 2] \cup {4} \cup (7, \infty)$

5. Absolute Value

For $x \in R$ , the absolute value (or modulus) is defined as:

$∣ x ∣ = {x, x \geq 0 - x, x < 0 .$

Equivalently, $∣ x ∣ = x^{2} = max {x, - x}$ . Geometrically, $∣ x ∣$ is the distance of $x$ from $0$ on the number line, and $∣ x - a ∣$ is the distance between $x$ and $a$ .

Figure: Absolute value y = |x| - the V-shaped graph with vertex at origin.

Basic Properties

$∣ - x ∣ = ∣ x ∣$
$∣ x y ∣ = ∣ x ∣ ∣ y ∣$ ; $\frac{x}{y} = \frac{∣ x ∣}{∣ y ∣}$ (for $y \neq = 0$ )
$∣ x ∣ > a ⟺ x > a$ or $x < - a$ , provided $a > 0$ ; if $a < 0$ , all real $x$ satisfy $∣ x ∣ > a$
$∣ x ∣ < a ⟺ - a < x < a$ , provided $a > 0$ ; no solution if $a \leq 0$
Triangle inequality: $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ , equality iff $x y \geq 0$
Reverse triangle inequality: $∣ x ∣ - ∣ y ∣ \leq ∣ x - y ∣$

Worked Example

Example

Show that

∣ x - 2∣ \leq 1 ⟺ 1 \leq x \leq 3

. (Three methods.)

Solution:

Method 1 - case-splitting.

Case (a): $x \geq 2 \Rightarrow x - 2 \leq 1 \Rightarrow x \leq 3$ . Combined: $2 \leq x \leq 3$ .

Case (b): $x < 2 \Rightarrow 2 - x \leq 1 \Rightarrow x \geq 1$ . Combined: $1 \leq x < 2$ .

Union: $x \in [1, 3]$ .

Method 2 - geometric. $∣ x - 2∣$ is the distance of $x$ from $2$ . So $∣ x - 2∣ \leq 1$ means $x$ is within $1$ unit of $2$ , i.e. $x \in [1, 3]$ .

Method 3 - squaring. Both sides non-negative, so square: $(x - 2)^{2} \leq 1 \Rightarrow x^{2} - 4 x + 3 \leq 0 \Rightarrow (x - 1) (x - 3) \leq 0 \Rightarrow x \in [1, 3]$ .

6. Logarithmic Function

The logarithm of $b$ to the base $a$ is the exponent to which $a$ must be raised to give $b$ :

$lo g_{a} b = x ⟺ a^{x} = b, a > 0, a \neq = 1, b > 0.$

Equivalently, the fundamental logarithmic identity:

$a^{l o g_{a} b} = b (a > 0, a \neq = 1, b > 0) .$

Graph of the Logarithmic Function

Figure: Logarithm

y = l o g_{a}

, x - increasing for a > 1, decreasing for 0 < a < 1; both pass through (1, 0).

Domain: $x > 0$ . Range: all real numbers. The graph passes through $(1, 0)$ for every base. The $y$ -axis is a vertical asymptote.

Properties of Logarithms

The expression $lo g_{a} b$ is meaningful only for $b > 0$ and either $0 < a < 1$ or $a > 1$ .

For $a > 1$ :

$lo g_{a} b = ⎩ ⎨ ⎧ > 0 if b > 1 0 if b = 1 < 0 if 0 < b < 1$

For $0 < a < 1$ :

$lo g_{a} b = ⎩ ⎨ ⎧ < 0 if b > 1 0 if b = 1 > 0 if 0 < b < 1$

Core algebra of logarithms:

$lo g_{a} (mn) = lo g_{a} m + lo g_{a} n$
$lo g_{a} (\frac{m}{n}) = lo g_{a} m - lo g_{a} n$
$lo g_{a} m^{n} = n lo g_{a} m$
$lo g_{a} 1 = 0$ ; $lo g_{a} a = 1$
Change of base: $lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}$ , valid for $c > 0$ , $c \neq = 1$
Reciprocal: $lo g_{b} a = \frac{1}{lo g _{a} b}$ , provided both $a$ and $b$ are positive and not $1$
Comparison: $lo g_{b} a_{1} \geq lo g_{b} a_{2} \Rightarrow a_{1} \geq a_{2}$ if $b > 1$ ; $a_{1} \leq a_{2}$ if $0 < b < 1$

Worked Example

Example

Solve for

x

4 lo g x = lo g (15 x^{2} + 16)

Solution:

$4 lo g x = lo g x^{4}$ , so $lo g x^{4} = lo g (15 x^{2} + 16)$ , giving $x^{4} = 15 x^{2} + 16$ .

Rearrange: $x^{4} - 15 x^{2} - 16 = 0$ , i.e. $(x^{2} - 16) (x^{2} + 1) = 0$ .

Since $x^{2} + 1 > 0$ , we get $x^{2} = 16 \Rightarrow x = \pm 4$ .

But $lo g x$ requires $x > 0$ , so $x = - 4$ is rejected. Answer: $x = 4$ .

Remark: Always test that the argument of

lo g

is positive after solving a log equation.

Solved Examples

Solved Example 1

a

and

b

are two rational numbers such that

(a^{2} - b + 1) + 3 (a^{2} + b - 3) = 0

, find

a

and

b

Solution:

Since $3$ is irrational and the sum of a rational and irrational number is zero only when both parts are zero, both the rational part and the coefficient of $3$ must vanish:

$a^{2} - b + 1 = 0$ and $a^{2} + b - 3 = 0$ .

Adding: $2 a^{2} - 2 = 0 \Rightarrow a^{2} = 1 \Rightarrow a = \pm 1$ . Substituting: $b = a^{2} + 1 = 2$ .

So $a = \pm 1$ , $b = 2$ .

Solved Example 2

Prove that

tan 5^{\circ}

is irrational.

Solution:

Assume, for contradiction, that $tan 5^{\circ}$ is rational. Then using the double-angle formula:

$tan 1 0^{\circ} = \frac{2 tan 5 ^{\circ}}{1 - tan ^{2} 5 ^{\circ}},$

which would also be rational (a ratio of rationals). By the triple-angle formula,

$tan 3 0^{\circ} = \frac{3 tan 1 0 ^{\circ} - tan ^{3} 1 0 ^{\circ}}{1 - 3 tan ^{2} 1 0 ^{\circ}},$

would then be rational too. But $tan 3 0^{\circ} = \frac{1}{3}$ is irrational. Contradiction.

Hence $tan 5^{\circ}$ is irrational.

Solved Example 3

Solve the inequality

x^{x - 2} > 1

Solution:

Case I: $x > 1$ . Since the base is greater than $1$ , $x^{x - 2} > 1 \Rightarrow x - 2 > 0 \Rightarrow x > 2$ . Solution: $x \in (2, \infty)$ .

Case II: $0 < x < 1$ . Since the base is between $0$ and $1$ , $x^{x - 2} > 1 \Rightarrow x - 2 < 0 \Rightarrow x < 2$ . Combined with case restriction: $x \in (0, 1)$ .

Combining: $x \in (0, 1) \cup (2, \infty)$ .

Solved Example 4

Find the maximum value of

\frac{2 x}{x ^{2} + 5 x + 1}

where

x \in R^{+}

Solution:

Let $y = \frac{2 x}{x ^{2} + 5 x + 1}$ . Dividing numerator and denominator by $x$ :

$y = \frac{2}{x + \frac{1}{x} + 5} .$

$y$ is maximum when the denominator is minimum. For $x > 0$ , by AM-GM, $x + \frac{1}{x} \geq 2$ , with equality at $x = 1$ . So the minimum denominator is $2 + 5 = 7$ .

Hence $y_{m a x} = \frac{2}{7}$ .

Solved Example 5

Solve

\frac{2 x + 3}{x ^{2} + x - 12} \leq \frac{1}{2}

Solution:

Move all terms to one side:

$\frac{2 x + 3}{x ^{2} + x - 12} - \frac{1}{2} \leq 0 \Rightarrow \frac{2 ( 2 x + 3 ) - ( x ^{2} + x - 12 )}{2 ( x ^{2} + x - 12 )} \leq 0 \Rightarrow \frac{- x ^{2} + 3 x + 18}{2 ( x ^{2} + x - 12 )} \leq 0.$

Multiply numerator and denominator by $- 1$ (flips inequality):

$\frac{x ^{2} - 3 x - 18}{2 ( x ^{2} + x - 12 )} \geq 0 \Rightarrow \frac{( x + 3 ) ( x - 6 )}{2 ( x + 4 ) ( x - 3 )} \geq 0.$

Critical points: $- 4$ (denominator, excluded), $- 3$ (numerator, included), $3$ (denominator, excluded), $6$ (numerator, included). All simple roots.

Sign check at $x = 7$ : all four factors positive, so expression $> 0$ . Alternating by the wavy curve method:

$x \in (- \infty, - 4) \cup [- 3, 3) \cup [6, \infty) .$

Solved Example 6

Solve the inequality

∣ x - 1∣ + ∣ x - 2∣ \leq 3

Solution:

Case I: $x \leq 1$ . Both expressions negate: $- (x - 1) - (x - 2) \leq 3 \Rightarrow - 2 x + 3 \leq 3 \Rightarrow x \geq 0$ . So $x \in [0, 1]$ .

Case II: $1 < x \leq 2$ . $(x - 1) - (x - 2) = 1 \leq 3$ . Always true. So $x \in (1, 2]$ .

Case III: $x > 2$ . $(x - 1) + (x - 2) \leq 3 \Rightarrow 2 x - 3 \leq 3 \Rightarrow x \leq 3$ . So $x \in (2, 3]$ .

Combining: $x \in [0, 3]$ .

Solved Example 7

Solve for real

x

:
(a)

∣ x - 3∣ > 5

(b)

∣2 x - 3∣ < 1

(c)

0 < ∣ x - 1∣ \leq 3

(d)

∣ x^{2} + x + 1∣ = x^{2} + x + 1

Solution:

(a) $∣ x - 3∣ > 5$ : distance from $3$ is more than $5$ , so $x < - 2$ or $x > 8$ . $x \in (- \infty, - 2) \cup (8, \infty)$ .

(b) $∣2 x - 3∣ < 1 \Rightarrow - 1 < 2 x - 3 < 1 \Rightarrow 2 < 2 x < 4 \Rightarrow 1 < x < 2$ . So $x \in (1, 2)$ .

(c) $0 < ∣ x - 1∣ \leq 3$ means $x \neq = 1$ and $- 3 \leq x - 1 \leq 3$ , i.e. $- 2 \leq x \leq 4$ . So $x \in [- 2, 4] - {1}$ .

(d) $∣ x^{2} + x + 1∣ = x^{2} + x + 1$ iff $x^{2} + x + 1 \geq 0$ . Since $x^{2} + x + 1 = (x + \frac{1}{2})^{2} + \frac{3}{4} > 0$ for all real $x$ , the equation holds for all $x \in R$ .

Solved Example 8

Solve for

x

lo g_{1/2} (x - 2) > 4

Solution:

Rewrite $4$ in the same base: $4 = lo g_{1/2} (\frac{1}{2})^{- 4} = lo g_{1/2} (\frac{1}{16})$ .

So the inequality becomes $lo g_{1/2} (x - 2) > lo g_{1/2} (\frac{1}{16})$ .

The base $\frac{1}{2} < 1$ , so the log function is decreasing; the inequality on arguments flips:

$x - 2 < \frac{1}{16} \Rightarrow x < \frac{33}{16} .$

Also the argument must be positive: $x - 2 > 0 \Rightarrow x > 2$ .

Combined: $x \in (2, \frac{33}{16})$ .

Solved Example 9

(I) Solve

lo g_{e} (x^{2} - 16) < lo g_{e} (4 x - 11)

. (II) If

lo g_{1/2} (x - 1) > 0

, find the interval of

x

Solution:

(I) The base $e > 1$ , so the inequality on logs gives the same inequality on arguments (provided both are positive):

$x^{2} - 16 < 4 x - 11 \Rightarrow x^{2} - 4 x - 5 < 0 \Rightarrow (x - 5) (x + 1) < 0 \Rightarrow x \in (- 1, 5)$ .

Also need $x^{2} - 16 > 0 \Rightarrow x < - 4$ or $x > 4$ , and $4 x - 11 > 0 \Rightarrow x > \frac{11}{4}$ .

Intersecting: $x \in (4, 5)$ .

(II) $lo g_{1/2} (x - 1) > 0 = lo g_{1/2} 1$ . Since the base $\frac{1}{2} < 1$ , the log function is decreasing, so the inequality flips on the arguments: $x - 1 < 1 \Rightarrow x < 2$ .

Also need $x - 1 > 0 \Rightarrow x > 1$ . Combined: $x \in (1, 2)$ .

Common Mistakes to Avoid

Watch out

Assuming $x^{2} = x$ . In fact $x^{2} = ∣ x ∣$ ; sign matters.
Multiplying an inequality by an expression whose sign you don't know. Multiplying by a negative flips the inequality; if the sign is unknown, split into cases or move everything to one side.
Cancelling $(x + 4)$ or $(x - 3)$ from both sides of a rational inequality. Cancellation loses the sign restrictions and often the solution.
Ignoring the domain of $lo g x$ . The argument must be strictly positive; solutions outside the domain must be rejected.
Forgetting the sign flip when the base of a log is between $0$ and $1$ . $lo g_{1/2} x$ is decreasing, so inequalities on the arguments flip direction.
In the wavy curve method, treating an even-multiplicity root the same as odd. At even roots the sign does not change; the curve touches and turns back.
Confusing $∣ x - a ∣$ with $x - a$ . The absolute value is always $\geq 0$ ; the expression inside can be negative.

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1. Number System

Natural Numbers (N)

Whole Numbers (W)

Integers (Z or I)

Rational Numbers (Q)

Irrational Numbers (Qc)

Real Numbers (R)

Complex Numbers (C)

2. Intervals

Finite Intervals

Infinite Intervals

3. Inequalities

Inequalities Involving Exponentials

4. Wavy Curve Method

The Method - Step by Step

Worked Example

5. Absolute Value

Basic Properties

Worked Example

6. Logarithmic Function

Graph of the Logarithmic Function

Properties of Logarithms

Worked Example

Solved Examples

Common Mistakes to Avoid

Ready to master Sets, Relations And Number System?

Natural Numbers ( $N$ )

Whole Numbers ( $W$ )

Integers ( $Z$ or $I$ )

Rational Numbers ( $Q$ )

Irrational Numbers ( $Q^{c}$ )

Real Numbers ( $R$ )

Complex Numbers ( $C$ )