CLASS XI | FOM - I

Fundamental of Mathematics – I

Number systems, notation, divisibility, unit digit cycles, prime factorisation, and sigma sums — with crisp examples and practice.

Contents

• Number system
• Decimal numbers
• Mathematical notation
• Types of numbers
• Prime factorisation, divisors, LCM & HCF
• Divisibility rules
• Unit place digit cycles
• Sigma notation and standard sums
• Question practice

Number system

Definition: A number system is a structured set of numbers along with operations and properties used to measure, count, and compare.

Natural numbers

\(\mathbb{N}=\{1,2,3,\dots\}\). Used for counting. No zero, no negatives.

Example: Successor of \(7\) is \(8\).

Practice:

1. Is \(0\) a natural number?
2. Find the 25th natural number.

Solutions

1. In this text, no. \(\mathbb{N}\) starts at 1 (some authors include 0).
2. \(25\).

Whole numbers

\(\mathbb{W}=\{0,1,2,3,\dots\}\). Natural numbers plus zero.

Example: \(0\in \mathbb{W}\) but \(0\notin \mathbb{N}\) (in this convention).

Practice: Which is the smallest whole number?

Solution

It is \(0\).

Integers

\(\mathbb{Z}=\{\dots,-3,-2,-1,0,1,2,3,\dots\}\). Includes negatives.

Example: Opposite of \(a\) is \(-a\).

Practice: Is \(-17\in\mathbb{Z}\)? Is \(-\frac12\in\mathbb{Z}\)?

Solution

\(-17\) yes; \(-\tfrac12\) no.

Rational numbers

\(\mathbb{Q}=\left\{ \frac{p}{q}\,:\, p,q\in\mathbb{Z},\, q\neq 0\right\}\). Fractions and integers.

Example: \(0.125=\frac{1}{8}\in\mathbb{Q}\).

Practice: Write \(2.3\overline{6}\) as \(p/q\).

Solution

\(2.3\overline{6}=\frac{71}{30}\).

Irrational numbers

\(\mathbb{R}\setminus\mathbb{Q}\). Cannot be written as \(p/q\). Non‑terminating, non‑repeating decimals.

Example: \(\sqrt{2},\, \pi,\, e\).

Practice: Classify \(0.101001000100001\ldots\).

Solution

Non‑terminating non‑repeating → irrational.

Real numbers

\(\mathbb{R}=\mathbb{Q}\cup(\mathbb{R}\setminus\mathbb{Q})\). All rationals and irrationals on the number line.

Practice: Is \(\sqrt{9}\in\mathbb{R}\)? Classify.

Solution

\(\sqrt{9}=3\in\mathbb{Q}\subset\mathbb{R}\).

Complex numbers

\(\mathbb{C}=\{a+bi: a,b\in\mathbb{R},\, i^2=-1\}\). Extends reals to solve all polynomials.

Example: Roots of \(x^2+1=0\) are \(x=\pm i\).

Practice: Write the real and imaginary parts of \(3-2i\).

Solution

Real \(=3\), Imag \(=-2\).

Decimal numbers

Decimals describe parts of a whole in base‑10. Patterns classify them into terminating and non‑terminating types.

Terminating decimals

Finite digits after the decimal point. Occur when the reduced denominator has only factors \(2\) and/or \(5\).

\(\frac{p}{q}\) (in lowest terms) is terminating ⇔ prime factors of \(q\) are only \(2\) or \(5\).

Example: \(\frac{7}{8}=0.875\) (denominator \(2^3\)).

Practice: Does \(\frac{3}{40}\) terminate?

Solution

Yes. \(40=2^3\cdot 5\).

Non‑terminating decimals

Infinite digits. Two kinds: repeating and non‑repeating.

Non‑terminating repeating

Digits form a repeating block; always rational.

Example: \(0.\overline{36}=\frac{4}{11}\).

Non‑terminating non‑repeating

No repeating pattern; always irrational.

Example: \(\pi=3.14159\ldots\).

\(p/q\) form

Every terminating or repeating decimal can be written as a fraction \(\frac{p}{q}\) with integers \(p,q\), \(q\neq 0\).

Practice: Convert \(0.2\overline{7}\) to \(p/q\).

Solution

\(x=0.2\overline{7}\Rightarrow 100x-10x=27\Rightarrow x=\frac{27}{90}=\frac{3}{10}\).

Mathematical notation

Universal quantifier

\(\forall x\in S,\, P(x)\) means “for all \(x\) in \(S\), \(P\) holds.”

Example: \(\forall n\in\mathbb{N},\; n(n+1)\) is even.

Existential quantifier

\(\exists x\in S\; \text{s.t.}\; P(x)\) means “there exists an \(x\) for which \(P\) holds.”

Example: \(\exists n\in\mathbb{Z}:\; n^2=4\) (namely \(n=\pm2\)).

Sigma notation

\(\displaystyle \sum_{k=m}^{n} a_k = a_m+a_{m+1}+\dots+a_n\).

Example: \(\sum_{k=1}^{5} k=15\).

Differentiation symbol

\(\frac{d}{dx}f(x)\) is the derivative; slope of the curve at \(x\).

Example: \(\frac{d}{dx}(x^3)=3x^2\).

Integration symbol

\(\displaystyle \int f(x)\,dx\) is the antiderivative; area accumulation.

Example: \(\int 3x^2\,dx = x^3+C\).

Practice:

1. Interpret \(\forall n\in\mathbb{N},\; n\ge 1\).
2. Expand \(\sum_{k=3}^{6} (2k-1)\).

Solutions

1. Every natural number is at least 1.
2. \(2\cdot3-1 + 2\cdot4-1 + 2\cdot5-1 + 2\cdot6-1 = 5+7+9+11=32\).

Types of numbers

Even and odd

Even: divisible by \(2\) (\(n=2k\)). Odd: of form \(2k+1\), not divisible by \(2\).

Example: Even ± even = even; odd ± odd = even; even × any = even.

Practice: Prove the sum of two consecutive integers is odd.

Solution

\(n+(n+1)=2n+1\) (odd form).

Prime and composite

Prime: >1 with exactly two divisors \(1\) and itself. Composite: >1 with more than two divisors.

Example: \(29\) is prime; \(28\) is composite.

Practice: Is \(1\) prime? composite? neither?

Solution

Neither; it has only one positive divisor.

Co‑prime (relatively prime)

Two integers \(a,b\) are co‑prime if \(\gcd(a,b)=1\) (they share no prime factor).

Example: \(8\) and \(15\) are co‑prime.

Practice: Are \(12\) and \(35\) co‑prime?

Solution

Yes, \(\gcd(12,35)=1\).

Prime factorisation, divisors, LCM & HCF

Factorisation into prime powers

Every \(n\ge 2\) can be uniquely written \(n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}\) with distinct primes \(p_i\) and exponents \(a_i\ge 1\).

Example: \(180=2^2\cdot 3^2\cdot 5^1\).

Total number of divisors

If \(n=\prod p_i^{a_i}\), then the count of positive divisors is \(\tau(n)=\prod (a_i+1)\).

Example: For \(180=2^2\cdot3^2\cdot5\), \(\tau(180)=(2+1)(2+1)(1+1)=18\).

LCM and HCF (GCD)

\[ \gcd\!\left(\prod p_i^{a_i},\; \prod p_i^{b_i}\right)=\prod p_i^{\min(a_i,b_i)},\quad \operatorname{lcm}\!\left(\prod p_i^{a_i},\; \prod p_i^{b_i}\right)=\prod p_i^{\max(a_i,b_i)} \]

Example: \(\gcd(60,90)=30\), \(\operatorname{lcm}(60,90)=180\).

Quick calculators

Practice:

1. Find the prime power form of \(252\) and the number of divisors.
2. Compute \(\gcd(84, 132)\) and \(\operatorname{lcm}(84,132)\).

Solutions

1. \(252=2^2\cdot 3^2\cdot 7\); divisors \((2+1)(2+1)(1+1)=18\).
2. \(\gcd=12\), \(\operatorname{lcm}=924\).

Divisibility rules

By 2

Last digit even (0,2,4,6,8).

By 3

Sum of digits divisible by \(3\).

By 4

Last two digits form a number divisible by \(4\).

By 5

Last digit \(0\) or \(5\).

By 6

Divisible by \(2\) and \(3\).

By 8

Last three digits form a number divisible by \(8\).

By 9

Sum of digits divisible by \(9\).

By 10

Last digit \(0\).

By 11

Difference of sums of alternating digits is a multiple of \(11\) (incl. \(0\)).

Composite numbers

Use prime factors: check divisibility for each prime power in the factorisation.

Divisibility quick check

Practice:

1. Test \(123456\) for divisibility by \(3, 4, 8, 9, 11\).
2. Find the smallest number to add to \(527\) to make it divisible by \(6\).

Solutions

1. \(3\): yes (sum=21); \(4\): last two 56 → yes; \(8\): last three 456 → yes; \(9\): 21 → no; \(11\): (1−2+3−4+5−6)=−3 → no.
2. Needs to be divisible by 2 and 3. Next even is 528 and sum 15 → divisible. Add \(1\).

Unit place digit cycles

The unit digit of \(a^n\) repeats in cycles mod \(10\). Knowing the cycle length gives the last digit quickly.

Cycles by base digit

Digit	Cycle (repeats)	Length
0	0	1
1	1	1
2	2,4,8,6	4
3	3,9,7,1	4
4	4,6	2
5	5	1
6	6	1
7	7,9,3,1	4
8	8,4,2,6	4
9	9,1	2

Unit digit finder

Practice:

1. Find the unit digit of \(17^{2025}\).
2. Find the unit digit of \(8^{73}\).

Solutions

1. Digit 7 cycle (7,9,3,1), \(2025\equiv 1\pmod 4\) → 7.
2. Digit 8 cycle (8,4,2,6), \(73\equiv 1\pmod 4\) → 8.

Sigma notation and standard sums

Symbol, limits, and expansion

\(\displaystyle \sum_{k=\alpha}^{\beta} f(k)\) adds values of \(f\) as \(k\) runs from lower limit \(\alpha\) to upper limit \(\beta\).

Expansion: \(\sum_{k=3}^{6} (k^2-1)=(3^2-1)+(4^2-1)+(5^2-1)+(6^2-1)\).

Forming sigma expressions

Repeated patterns compress into \(\sum\). For odd numbers \(1,3,5,\dots,(2n-1)\): \(\sum_{k=1}^n (2k-1)\).

Standard results

Sum of first \(n\) natural numbers
\[\sum_{k=1}^{n} k=\frac{n(n+1)}{2}\]

Sum of squares
\[\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}\]

Sum of cubes
\[\sum_{k=1}^{n} k^3=\left(\frac{n(n+1)}{2}\right)^2\]

Sum of a constant
\[\sum_{k=1}^{n} c = c\cdot n\]

Quick sigma calculator (standard sums)

∑ k ∑ k² ∑ k³ ∑ constant c

Practice:

1. Evaluate \(\sum_{k=1}^{50} k\).
2. Evaluate \(\sum_{k=1}^{20} (3k-2)\).

Solutions

1. \(\frac{50\cdot 51}{2}=1275\).
2. \(3\sum k - 2\cdot 20 = 3\cdot \frac{20\cdot 21}{2}-40=590\).

Question practice

Mixed practice across all subtopics. Use the calculators above only for verification—try reasoning first.

1. Number system: Classify each as rational/irrational: \(\sqrt{50}\), \(0.\overline{142857}\), \(0.1010010001\ldots\).
2. Decimals: Decide if \(\frac{77}{120}\) is terminating. If yes, find the number of decimal places.
3. Notation: Write “sum of first \(n\) odd numbers” using sigma and evaluate it.
4. Types: Find the smallest composite that is a multiple of both \(12\) and \(15\).
5. Prime factors: For \(n=1320\), find its prime power form, \(\tau(n)\), and number of odd divisors.
6. Divisibility: Determine whether \(7{,}654{,}321\) is divisible by \(3\), \(9\), or \(11\).
7. Unit digit: Find the unit digit of \(3^{2024}+7^{2025}\).
8. Sigma: Compute \(\sum_{k=1}^{30} (k^2+k)\).

Answer key

1. \(\sqrt{50}=5\sqrt{2}\) irrational; \(0.\overline{142857}\) rational; \(0.1010010001\ldots\) irrational.
2. \(120=2^3\cdot 3\cdot 5\) → has a factor \(3\) → non‑terminating repeating.
3. \(\sum_{k=1}^{n} (2k-1)=n^2\).
4. \(\operatorname{lcm}(12,15)=60\).
5. \(1320=2^3\cdot 3\cdot 5\cdot 11\). \(\tau= (3+1)(1+1)(1+1)(1+1)=32\). Odd divisors: ignore \(2^3\) → \(2\cdot2\cdot2=8\).
6. Sum of digits \(=34\) → not divisible by \(3\) or \(9\). For \(11\): alternating sum \(=(7−6+5−4+3−2+1)=4\) → not divisible.
7. Unit( \(3^{2024}\) ) cycle (3,9,7,1) with \(2024\equiv 0\pmod4\) → 1. Unit( \(7^{2025}\) ) cycle (7,9,3,1) with \(2025\equiv 1\pmod4\) → 7. Total unit digit \(=1+7=8\).
8. \(\sum k^2+\sum k=\frac{30\cdot31\cdot61}{6}+\frac{30\cdot31}{2}= (30\cdot31)\left(\frac{61}{6}+\frac{1}{2}\right)=(930)\cdot\frac{64}{6}=930\cdot\frac{32}{3}=9920\).

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