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Absolute Value and Integer Operations

Integer Operations

Integers are the positive and negative whole numbers, including zero:

{…,−3,−2,−1,0,1,2,3,… }\{\dots,-3,-2,-1,0,1,2,3,\dots\}

Addition and Subtraction

  • Same signs: add and keep the sign.
  • Different signs: subtract and keep the sign of the larger absolute value.

For example:

−7+3=−4-7 + 3 = -4

Multiplication and Division

  • Same signs give a positive result.
  • Different signs give a negative result.

For example:

(−4)(−5)=20(-4)(-5)=20
(−4)(5)=−20(-4)(5)=-20

Absolute Value

The absolute value of a number is its distance from 00 on the number line.

Absolute value is written using vertical bars:

∣x∣|x|

For example:

∣5∣=5|5| = 5

and:

∣−5∣=5|-5| = 5

because both 55 and −5-5 are 55 units away from 00.

In general:

∣x∣={x,x≥0−x,x<0|x|= \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}

Properties of Absolute Value

Product Rule

∣ab∣=∣a∣∣b∣|ab| = |a||b|

Example:

∣−3⋅4∣=∣−12∣=12|-3 \cdot 4| = |-12| = 12

and:

∣−3∣∣4∣=3⋅4=12|-3||4| = 3 \cdot 4 = 12

Quotient Rule

∣ab∣=∣a∣∣b∣(b≠0)\left|\frac{a}{b}\right| = \frac{|a|}{|b|} \quad (b \ne 0)

Triangle Inequality

∣a+b∣≤∣a∣+∣b∣|a+b| \le |a| + |b|

This means the distance traveled together cannot exceed the sum of the individual distances.

Example 1:

Find all solutions to the equation |x ^ 2 - 3x| = 4 .

Solution: We can split the problem into two cases: either

x2−3x=4 or x2−3x=−4x ^ 2 - 3x = 4 \\ \text{ or } \\ x ^ 2 - 3x = - 4

The first case yields the solutions 4 and -1; the second yields 3±i72\frac{3 \pm i \sqrt{7}}{2} If we restrict ourselves to real solutions, only 4 and -1 are acceptable.

Example 2: Solve the equation |(x + 2) / (3x - 1)| = 5

This gives two cases:

x+23x−1=5orx+23x−1=−5\frac{x+2}{3x-1}=5 \quad \text{or} \quad \frac{x+2}{3x-1}=-5

For the first case:

x+2=5(3x−1)x+2=5(3x-1)
x+2=15x−5x+2=15x-5
7=14x7=14x
x=12x=\frac12

For the second case:

x+2=−5(3x−1)x+2=-5(3x-1)
x+2=−15x+5x+2=-15x+5
16x=316x=3
x=316x=\frac{3}{16}

Since:

3x−1≠03x-1 \ne 0

we must have:

x≠13x \ne \frac13

Both solutions are valid.

Thus, the solutions are:

x=12 or x=316\boxed{x=\frac12 \text{ or } x=\frac{3}{16}}

Sigma Notation

Sigma notation represents sums.

The Greek letter sigma is written as:

∑\sum

For example:

∑k=15k\sum_{k=1}^{5} k

means:

1+2+3+4+51+2+3+4+5

which equals:

1515

Example

∑n=14n2\sum_{n=1}^{4} n^2

means:

12+22+32+421^2+2^2+3^2+4^2

which equals:

3030

Important Summation Formulae

Sum of the First (n) Natural Numbers

∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}
Proof

Let:

S=1+2+3+⋯+nS = 1+2+3+\dots+n

Write the sum in reverse order:

S=n+(n−1)+(n−2)+⋯+1S = n+(n-1)+(n-2)+\dots+1

Adding the two equations term by term:

2S=(n+1)+(n+1)+⋯+(n+1)2S=(n+1)+(n+1)+\dots+(n+1)

There are (n) terms, so:

2S=n(n+1)2S=n(n+1)

Thus:

S=n(n+1)2S=\frac{n(n+1)}{2}

Sum of the Squares of the First (n) Natural Numbers

∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
Proof Sketch

One way to prove this formula is by mathematical induction.

Assume:

∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

Then:

∑k=1n+1k2=n(n+1)(2n+1)6+(n+1)2\sum_{k=1}^{n+1} k^2 = \frac{n(n+1)(2n+1)}{6}+(n+1)^2

Factoring and simplifying gives:

(n+1)(n+2)(2n+3)6\frac{(n+1)(n+2)(2n+3)}{6}

which matches the formula with (n+1).

Thus, the formula holds for all positive integers (n).

Sum of the Cubes of the First (n) Natural Numbers

∑k=1nk3=(n(n+1)2)2\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2
Interesting Observation

The sum of cubes equals the square of the sum of the first (n) natural numbers.

That is:

13+23+⋯+n3=(1+2+⋯+n)21^3+2^3+\dots+n^3 = (1+2+\dots+n)^2

For example:

13+23+33=1+8+27=361^3+2^3+3^3 = 1+8+27 = 36

and:

(1+2+3)2=62=36(1+2+3)^2 = 6^2 = 36

This identity can also be proved using induction.

Pi Notation

Pi notation represents products.

The Greek capital pi is written as:

∏\prod

For example:

∏k=14k\prod_{k=1}^{4} k

means:

1⋅2⋅3⋅41 \cdot 2 \cdot 3 \cdot 4

which equals:

2424

Another example:

∏n=25n\prod_{n=2}^{5} n

means:

2⋅3⋅4⋅52 \cdot 3 \cdot 4 \cdot 5

which equals:

120120

Part 2

Great work if you read all the way till here! Due to File-Size Constraints, this module has been divided into 2 parts. Continue this Module in part 2. Good Luck!

Module Progress:

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