Real Numbers Notes

1

Real Numbers

Detailed Theory • Very Easy English • Board Exam Focus

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Chapter Roadmap

1. Number System 2. Euclid's Division Lemma 3. Euclid's Division Algorithm 4. HCF Using Euclid's Algorithm 5. Fundamental Theorem of Arithmetic 6. HCF and LCM by Prime Factorisation 7. Irrational Numbers 8. Proof of Irrationality 9. Decimal Expansions 10. Extra Board Questions 11. Final Revision

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1. Real Numbers and the Number System

Natural Numbers

Natural numbers are the counting numbers. They are usually written as \(1,2,3,4,\ldots\).

\[\boxed{\mathbb N=\{1,2,3,4,\ldots\}}\]

Whole Numbers

Whole numbers include \(0\) together with all natural numbers.

\[\boxed{\mathbb W=\{0,1,2,3,4,\ldots\}}\]

Integers

Integers include positive numbers, negative numbers and zero.

\[\boxed{\mathbb Z=\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}}\]

Rational Numbers

A number is rational if it can be written in the form

\[ \boxed{\frac pq,\qquad p,q\in\mathbb Z,\quad q\neq0} \]

Examples are \(\frac23,\;-5,\;0,\;1.25\).

Irrational Numbers

A number that cannot be written in the form \(p/q\), where \(p,q\) are integers and \(q\neq0\), is called an irrational number.

Examples include \(\sqrt2,\sqrt3,\sqrt5,\pi\).

Real Numbers

All rational and irrational numbers together form the set of real numbers.

\[ \boxed{\text{Real Numbers}=\text{Rational Numbers}\cup\text{Irrational Numbers}} \]

Three Detailed Solved Questions

Question 1. Classify \(0,\;-7,\;\frac35,\;\sqrt2\) as rational or irrational.

Solution:

1. \(0\) can be written as \(\frac01\), where both \(0\) and \(1\) are integers and denominator is non-zero.
2. Therefore, \(0\) is rational.
3. \(-7\) can be written as \(\frac{-7}{1}\).
4. Therefore, \(-7\) is rational.
5. \(\frac35\) is already in the form \(p/q\), with \(3,5\) integers and \(5\neq0\).
6. Therefore, \(\frac35\) is rational.
7. \(\sqrt2\) cannot be expressed as a ratio of two integers.
8. Therefore, \(\sqrt2\) is irrational.

Answer: \(0,-7,\frac35\) are rational; \(\sqrt2\) is irrational.

Question 2. Is \(2.75\) a rational number?

Solution:

1. Given decimal number is \(2.75\).
2. There are two digits after the decimal point.
3. Remove the decimal point and write the number over \(100\): \[ 2.75=\frac{275}{100} \]
4. Divide numerator and denominator by \(25\): \[ \frac{275}{100}=\frac{11}{4} \]
5. \(11\) and \(4\) are integers and \(4\neq0\).

Answer: Yes, \(2.75\) is rational.

Question 3. Explain why every integer is a rational number.

Solution:

1. Let \(n\) be any integer.
2. We can write \(n\) as: \[ n=\frac n1 \]
3. Here \(n\) is an integer and \(1\) is also an integer.
4. The denominator \(1\) is not zero.
5. Therefore, \(n\) satisfies the form \(p/q\), where \(p,q\in\mathbb Z\) and \(q\neq0\).

Answer: Every integer is a rational number.

2. Euclid's Division Lemma

Statement

For given positive integers \(a\) and \(b\), there exist unique integers \(q\) and \(r\) such that

\[ \boxed{a=bq+r} \] \[ \boxed{0\le r

Here:

• \(a\) = dividend
• \(b\) = divisor
• \(q\) = quotient
• \(r\) = remainder

Most Important Condition: The remainder must always satisfy \(0\le r

Three Detailed Solved Questions

Question 1. Write \(37\) in the form \(bq+r\), where \(b=5\).

Solution:

1. We have dividend \(a=37\) and divisor \(b=5\).
2. Divide \(37\) by \(5\).
3. \(5\) goes into \(37\) seven times: \[ 5\times7=35 \]
4. Subtract: \[ 37-35=2 \]
5. Therefore, quotient \(q=7\) and remainder \(r=2\).
6. Using \(a=bq+r\): \[ 37=5(7)+2 \]
7. Check the condition: \[ 0\le2<5 \] which is true.

Answer: \(\boxed{37=5(7)+2}\).

Question 2. Divide \(125\) by \(8\) and verify Euclid's division lemma.

Solution:

1. Dividend \(a=125\), divisor \(b=8\).
2. Divide \(125\) by \(8\).
3. \(8\times15=120\).
4. Remainder: \[ 125-120=5 \]
5. Therefore \(q=15,r=5\).
6. Write: \[ 125=8(15)+5 \]
7. Check: \[ 0\le5<8 \]
8. The condition is satisfied.

Answer: \(125=8(15)+5\), so the lemma is verified.

Question 3. Can \(17=4(3)+5\) represent Euclid's division lemma?

Solution:

1. Compare with \(a=bq+r\).
2. Here \(a=17,b=4,q=3,r=5\).
3. The first condition is satisfied because: \[ 17=4(3)+5 \]
4. Now check the remainder condition: \[ 0\le r
5. Here: \[ 0\le5<4 \]
6. But \(5<4\) is false.
7. Therefore, the required condition is not satisfied.

Answer: No, this is not a valid Euclidean division form for divisor \(4\).

3. Euclid's Division Algorithm

Purpose

Euclid's division algorithm is a systematic method used to find the HCF (Highest Common Factor) of two positive integers.

Method

Suppose \(a>b\). Divide \(a\) by \(b\):

\[ a=bq+r \]

If \(r\neq0\), divide \(b\) by \(r\). Continue this process until the remainder becomes \(0\).

The divisor of the last division, i.e. the last non-zero remainder, is the HCF.

Board Formula:

\[ \boxed{\mathrm{HCF}(a,b)=\mathrm{HCF}(b,r)} \]

Three Detailed Solved Questions

Question 1. Find the HCF of \(135\) and \(225\) using Euclid's division algorithm.

Solution:

1. Since \(225>135\), divide \(225\) by \(135\).
2. \[ 225=135(1)+90 \]
3. Now divide \(135\) by \(90\): \[ 135=90(1)+45 \]
4. Now divide \(90\) by \(45\): \[ 90=45(2)+0 \]
5. The remainder is now \(0\).
6. The last non-zero remainder is \(45\).

Answer: \(\boxed{\mathrm{HCF}(135,225)=45}\).

Question 2. Find the HCF of \(867\) and \(255\) using Euclid's algorithm.

Solution:

1. Since \(867>255\), divide \(867\) by \(255\): \[ 867=255(3)+102 \]
2. Now divide \(255\) by \(102\): \[ 255=102(2)+51 \]
3. Now divide \(102\) by \(51\): \[ 102=51(2)+0 \]
4. The last non-zero remainder is \(51\).

Answer: \(\boxed{51}\).

Question 3. Find the HCF of \(196\) and \(38220\).

Solution:

1. Take the larger number \(38220\) and divide by \(196\): \[ 38220=196(195)+0 \]
2. The remainder is already \(0\).
3. Therefore, the divisor \(196\) is the last non-zero remainder.

Answer: \(\boxed{\mathrm{HCF}=196}\).

4. HCF Using Euclid's Algorithm – Board Method

Why Does It Work?

If

\[ a=bq+r \]

then every common divisor of \(a\) and \(b\) also divides \(r=a-bq\). Therefore, the common divisors of \(a,b\) are the same as the common divisors of \(b,r\). This gives:

\[ \boxed{\mathrm{HCF}(a,b)=\mathrm{HCF}(b,r)} \]

Three Detailed Solved Questions

Question 1. Find the HCF of \(616\) and \(32\).

Solution:

1. Divide \(616\) by \(32\): \[ 616=32(19)+8 \]
2. Now divide \(32\) by \(8\): \[ 32=8(4)+0 \]
3. The last non-zero remainder is \(8\).

Answer: \(\boxed{8}\).

Question 2. Show that the HCF of \(306\) and \(657\) is \(9\).

Solution:

1. Divide \(657\) by \(306\): \[ 657=306(2)+45 \]
2. Divide \(306\) by \(45\): \[ 306=45(6)+36 \]
3. Divide \(45\) by \(36\): \[ 45=36(1)+9 \]
4. Divide \(36\) by \(9\): \[ 36=9(4)+0 \]
5. The last non-zero remainder is \(9\).

Answer: \(\boxed{\mathrm{HCF}=9}\).

Question 3. Find the greatest number that divides \(245\) and \(1029\), leaving remainders \(5\) and \(9\), respectively.

Solution:

1. Let the required number be \(d\).
2. Since the remainder is \(5\) from \(245\), \(d\) divides: \[ 245-5=240 \]
3. Since the remainder is \(9\) from \(1029\), \(d\) divides: \[ 1029-9=1020 \]
4. Therefore, \(d\) must be the HCF of \(240\) and \(1020\).
5. Apply Euclid's algorithm: \[ 1020=240(4)+60 \]
6. Then: \[ 240=60(4)+0 \]
7. The HCF is \(60\).

Answer: The greatest required number is \(\boxed{60}\).

5. Fundamental Theorem of Arithmetic

Statement

The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of the prime factors.

\[ \boxed{\text{Composite number}=\text{product of prime numbers}} \]

Example

\[ 60=2\times2\times3\times5=2^2\times3\times5 \]

This prime factorisation is unique apart from the order of the factors.

Three Detailed Solved Questions

Question 1. Find the prime factorisation of \(360\).

Solution:

1. Divide \(360\) by the smallest prime \(2\): \[ 360=2\times180 \]
2. Again divide \(180\) by \(2\): \[ 180=2\times90 \]
3. Again: \[ 90=2\times45 \]
4. Now \(45\) is divisible by \(3\): \[ 45=3\times15 \]
5. Again: \[ 15=3\times5 \]
6. \(5\) is prime.
7. Therefore: \[ 360=2\times2\times2\times3\times3\times5 \]
8. Using powers: \[ \boxed{360=2^3\times3^2\times5} \]

Question 2. Find the prime factorisation of \(1575\).

Solution:

1. \(1575\) is divisible by \(3\), because \(1+5+7+5=18\).
2. \[ 1575=3\times525 \]
3. Again: \[ 525=3\times175 \]
4. Now: \[ 175=5\times35 \]
5. And: \[ 35=5\times7 \]
6. \(7\) is prime.
7. Therefore: \[ 1575=3\times3\times5\times5\times7 \]
8. Hence: \[ \boxed{1575=3^2\times5^2\times7} \]

Question 3. Show that \(2^3\times3^2\times5\) is the prime factorisation of \(360\).

Solution:

1. Expand the powers: \[ 2^3=2\times2\times2 \]
2. Also: \[ 3^2=3\times3 \]
3. Therefore: \[ 2^3\times3^2\times5 =2\times2\times2\times3\times3\times5 \]
4. Multiply: \[ 2\times2\times2=8 \]
5. \[ 3\times3=9 \]
6. Then: \[ 8\times9\times5=72\times5=360 \]

Answer: The prime factorisation is correct.

6. HCF and LCM by Prime Factorisation

Rule for HCF

Write each number as a product of primes. For HCF, take the smallest power of every prime common to all numbers.

Rule for LCM

For LCM, take the greatest power of every prime appearing in the factorisations.

\[ \boxed{\text{HCF: smallest powers of common primes}} \] \[ \boxed{\text{LCM: greatest powers of all primes}} \]

Relationship Between HCF and LCM

For two positive integers \(a\) and \(b\):

\[ \boxed{\mathrm{HCF}(a,b)\times\mathrm{LCM}(a,b)=a\times b} \]

Three Detailed Solved Questions

Question 1. Find the HCF and LCM of \(72\) and \(120\) by prime factorisation.

Solution:

1. Prime factorise \(72\): \[ 72=2^3\times3^2 \]
2. Prime factorise \(120\): \[ 120=2^3\times3\times5 \]
3. For HCF, take common primes with smallest powers: \[ \mathrm{HCF}=2^3\times3 \]
4. \[ =8\times3=24 \]
5. For LCM, take all primes with greatest powers: \[ \mathrm{LCM}=2^3\times3^2\times5 \]
6. \[ =8\times9\times5=360 \]
7. Verify: \[ 24\times360=8640 \]
8. Also: \[ 72\times120=8640 \]

Answer: HCF \(=\boxed{24}\), LCM \(=\boxed{360}\).

Question 2. Find the HCF and LCM of \(84\) and \(140\).

Solution:

1. Prime factorise \(84\): \[ 84=2^2\times3\times7 \]
2. Prime factorise \(140\): \[ 140=2^2\times5\times7 \]
3. Common primes are \(2\) and \(7\).
4. Smallest powers give: \[ \mathrm{HCF}=2^2\times7=4\times7=28 \]
5. For LCM, use greatest powers of \(2,3,5,7\): \[ \mathrm{LCM}=2^2\times3\times5\times7 \]
6. \[ =4\times3\times5\times7=420 \]

Answer: HCF \(=\boxed{28}\), LCM \(=\boxed{420}\).

Question 3. The HCF of two numbers is \(12\), their LCM is \(420\), and one number is \(60\). Find the other number.

Solution:

1. Let the other number be \(x\).
2. Use: \[ \mathrm{HCF}\times\mathrm{LCM}=\text{product of numbers} \]
3. Substitute the given values: \[ 12\times420=60x \]
4. Multiply: \[ 5040=60x \]
5. Divide both sides by \(60\): \[ x=\frac{5040}{60} \]
6. \[ x=84 \]

Answer: The other number is \(\boxed{84}\).

7. Irrational Numbers – Detailed Concept

Meaning

An irrational number cannot be expressed as \(\frac pq\), where \(p,q\) are integers and \(q\neq0\).

Important Examples

• \(\sqrt2\)
• \(\sqrt3\)
• \(\sqrt5\)
• \(\pi\)

Important: Not every square root is irrational. For example, \(\sqrt9=3\), which is rational. A square root of a non-perfect square is irrational.

Three Detailed Solved Questions

Question 1. Is \(\sqrt{49}\) rational or irrational?

Solution:

1. Calculate the square root: \[ \sqrt{49}=7 \]
2. \(7\) is an integer.
3. Every integer is rational because \(7=\frac71\).

Answer: \(\sqrt{49}\) is rational.

Question 2. Is \(\sqrt7\) rational or irrational?

Solution:

1. \(7\) is not a perfect square.
2. Therefore, \(\sqrt7\) cannot be represented as an integer.
3. In fact, \(\sqrt7\) cannot be expressed as a ratio of two integers.

Answer: \(\sqrt7\) is irrational.

Question 3. Classify \(\frac{\sqrt{16}}{3}\).

Solution:

1. Evaluate the square root: \[ \sqrt{16}=4 \]
2. Therefore: \[ \frac{\sqrt{16}}3=\frac43 \]
3. \(4\) and \(3\) are integers and \(3\neq0\).
4. Hence the number is of the form \(p/q\).

Answer: \(\frac{\sqrt{16}}3\) is rational.

8. Proving That a Number Is Irrational

Standard Board-Exam Proof Method

To prove a number such as \(\sqrt2\) is irrational, use contradiction.

1. Assume the number is rational.
2. Write it in lowest form as \(\frac pq\), where \(p,q\) are integers, \(q\neq0\), and \(\gcd(p,q)=1\).
3. Square or manipulate the equation.
4. Show that both \(p\) and \(q\) must have a common factor.
5. This contradicts \(\gcd(p,q)=1\).
6. Therefore, the original number is irrational.

Three Detailed Solved Questions

Question 1. Prove that \(\sqrt2\) is irrational.

Proof:

1. Assume, to the contrary, that \(\sqrt2\) is rational.
2. Therefore, it can be written as: \[ \sqrt2=\frac pq \] where \(p,q\) are integers, \(q\neq0\), and \(p,q\) have no common factor.
3. Square both sides: \[ 2=\frac{p^2}{q^2} \]
4. Multiply by \(q^2\): \[ p^2=2q^2 \]
5. Therefore, \(p^2\) is divisible by \(2\).
6. Hence \(p\) is divisible by \(2\).
7. Let \(p=2k\), where \(k\) is an integer.
8. Substitute: \[ (2k)^2=2q^2 \]
9. \[ 4k^2=2q^2 \]
10. Divide by \(2\): \[ 2k^2=q^2 \]
11. Thus \(q^2\) is divisible by \(2\), so \(q\) is also divisible by \(2\).
12. Therefore, both \(p\) and \(q\) are divisible by \(2\).
13. This contradicts the fact that \(p\) and \(q\) have no common factor.

Hence, \(\boxed{\sqrt2\text{ is irrational}}\).

Question 2. Prove that \(\sqrt3\) is irrational.

Proof:

1. Assume that \(\sqrt3\) is rational.
2. Then: \[ \sqrt3=\frac pq \] where \(p,q\) are coprime integers and \(q\neq0\).
3. Squaring: \[ 3=\frac{p^2}{q^2} \]
4. Therefore: \[ p^2=3q^2 \]
5. Hence \(p^2\) is divisible by \(3\).
6. So \(p\) is divisible by \(3\).
7. Let \(p=3k\).
8. Substitute: \[ (3k)^2=3q^2 \]
9. \[ 9k^2=3q^2 \]
10. Divide by \(3\): \[ 3k^2=q^2 \]
11. Thus \(q^2\) is divisible by \(3\), so \(q\) is divisible by \(3\).
12. Therefore both \(p\) and \(q\) are divisible by \(3\), a contradiction.

Hence, \(\boxed{\sqrt3\text{ is irrational}}\).

Question 3. Prove that \(5+\sqrt2\) is irrational.

Proof:

1. Assume that \(5+\sqrt2\) is rational.
2. Since \(5\) is rational, subtract \(5\) from both sides.
3. If \(5+\sqrt2\) were rational, then: \[ (5+\sqrt2)-5=\sqrt2 \] would also be rational.
4. But \(\sqrt2\) is irrational.
5. This is a contradiction.

Hence, \(\boxed{5+\sqrt2\text{ is irrational}}\).

9. Decimal Expansions of Rational Numbers

Terminating Decimal

A rational number has a terminating decimal expansion if its decimal representation ends after a finite number of digits.

For a rational number \(\frac pq\) in lowest form, its decimal expansion terminates if the prime factorisation of \(q\) is of the form:

\[ \boxed{q=2^m5^n} \] where \(m,n\) are non-negative integers.

Non-Terminating Recurring Decimal

If, after simplification, the denominator contains any prime factor other than \(2\) or \(5\), the decimal expansion is non-terminating recurring.

\[ \boxed{\text{Denominator contains only }2,5\Rightarrow\text{ terminating}} \] \[ \boxed{\text{Any other prime factor}\Rightarrow\text{ non-terminating recurring}} \]

Three Detailed Solved Questions

Question 1. Determine whether \(\frac7{40}\) has a terminating decimal expansion.

Solution:

1. The fraction is already in lowest form because \(7\) and \(40\) have no common factor.
2. Prime factorise the denominator: \[ 40=2^3\times5 \]
3. The denominator contains only the prime factors \(2\) and \(5\).
4. Therefore, the decimal expansion terminates.
5. Now convert it: \[ \frac7{40}=\frac{7\times25}{40\times25} =\frac{175}{1000} \]
6. Therefore: \[ \frac7{40}=0.175 \]

Answer: Terminating decimal, \(\boxed{0.175}\).

Question 2. Determine whether \(\frac{13}{24}\) has a terminating decimal expansion.

Solution:

1. The fraction is in lowest form because \(\gcd(13,24)=1\).
2. Prime factorise \(24\): \[ 24=2^3\times3 \]
3. The denominator contains \(3\), which is neither \(2\) nor \(5\).
4. Therefore, the decimal expansion does not terminate.
5. It is a non-terminating recurring decimal.

Answer: \(\boxed{\frac{13}{24}}\) has a non-terminating recurring decimal expansion.

Question 3. Determine whether \(\frac{21}{150}\) has a terminating decimal expansion.

Solution:

1. First reduce the fraction to lowest form.
2. Divide numerator and denominator by \(3\): \[ \frac{21}{150}=\frac7{50} \]
3. Prime factorise \(50\): \[ 50=2\times5^2 \]
4. The denominator contains only \(2\) and \(5\).
5. Therefore, the decimal expansion terminates.
6. Convert: \[ \frac7{50}=\frac{14}{100}=0.14 \]

Answer: Terminating decimal, \(\boxed{0.14}\).

10. Decimal Expansion – Important Board Concepts

Why Must We Reduce the Fraction First?

The denominator test is applied to the denominator of the rational number in lowest form. A factor \(3\), for example, may disappear after cancellation.

Consider:

\[ \frac{30}{75}=\frac25 \]

The original denominator \(75\) contains \(3\), but after simplification the denominator is \(5\). Therefore the decimal terminates.

Three Detailed Solved Questions

Question 1. Without actual division, determine whether \(\frac{17}{125}\) has a terminating decimal expansion.

Solution:

1. The denominator is \(125\).
2. Prime factorise: \[ 125=5^3 \]
3. Only the prime factor \(5\) occurs.
4. Therefore the decimal expansion terminates.

Answer: Terminating decimal expansion.

Question 2. Without actual division, determine whether \(\frac{29}{2^4\times5^3}\) terminates.

Solution:

1. The denominator is already expressed as: \[ 2^4\times5^3 \]
2. The only prime factors are \(2\) and \(5\).
3. Therefore the denominator has the required form \(2^m5^n\).

Answer: The decimal expansion terminates.

Question 3. Determine whether \(\frac{7}{2^3\times3\times5}\) terminates.

Solution:

1. The denominator is: \[ 2^3\times3\times5 \]
2. The prime factors are \(2,3,5\).
3. The factor \(3\) is present.
4. The denominator is therefore not of the form \(2^m5^n\).

Answer: The decimal expansion is non-terminating recurring.

11. 10 Extra Fully Detailed Board-Style Questions

Question 1. Use Euclid's division algorithm to find the HCF of \(135\) and \(225\).

1. \(225=135(1)+90\)
2. \(135=90(1)+45\)
3. \(90=45(2)+0\)
4. The last non-zero remainder is \(45\).

HCF \(=\boxed{45}\).

Question 2. Find the HCF and LCM of \(18\) and \(24\) by prime factorisation.

1. \[ 18=2\times3^2 \]
2. \[ 24=2^3\times3 \]
3. HCF uses smallest powers: \[ 2^1\times3^1=6 \]
4. LCM uses greatest powers: \[ 2^3\times3^2=8\times9=72 \]
5. Check: \[ 6\times72=432 \]
6. Also: \[ 18\times24=432 \]

HCF \(=\boxed6\), LCM \(=\boxed{72}\).

Question 3. Prove that \(\sqrt5\) is irrational.

1. Assume \(\sqrt5\) is rational.
2. Write: \[ \sqrt5=\frac pq \] where \(p,q\) are coprime integers and \(q\neq0\).
3. Squaring: \[ 5=\frac{p^2}{q^2} \]
4. Thus: \[ p^2=5q^2 \]
5. So \(5\mid p^2\), hence \(5\mid p\).
6. Let \(p=5k\).
7. Then: \[ 25k^2=5q^2 \]
8. Therefore: \[ q^2=5k^2 \]
9. So \(5\mid q\).
10. Thus \(5\) divides both \(p\) and \(q\), contradicting their coprimality.

Hence \(\boxed{\sqrt5\text{ is irrational}}\).

Question 4. Find the prime factorisation of \(540\).

1. \[ 540=2\times270 \]
2. \[ 270=2\times135 \]
3. \[ 135=3\times45 \]
4. \[ 45=3\times15 \]
5. \[ 15=3\times5 \]
6. Therefore: \[ 540=2\times2\times3\times3\times3\times5 \]
7. Hence: \[ \boxed{540=2^2\times3^3\times5} \]

Question 5. Find the least number which is divisible by \(12,18\) and \(30\).

1. We need the LCM of \(12,18,30\).
2. Prime factorise: \[ 12=2^2\times3 \]
3. \[ 18=2\times3^2 \]
4. \[ 30=2\times3\times5 \]
5. Take greatest power of every prime: \[ \mathrm{LCM}=2^2\times3^2\times5 \]
6. \[ =4\times9\times5 \]
7. \[ =36\times5=180 \]

Answer: \(\boxed{180}\).

Question 6. Find the greatest number that divides \(398\) and \(436\), leaving remainder \(8\) in each case.

1. Subtract the common remainder \(8\) from both numbers.
2. \[ 398-8=390 \]
3. \[ 436-8=428 \]
4. The required number is the HCF of \(390\) and \(428\).
5. Use Euclid: \[ 428=390(1)+38 \]
6. \[ 390=38(10)+10 \]
7. \[ 38=10(3)+8 \]
8. \[ 10=8(1)+2 \]
9. \[ 8=2(4)+0 \]
10. Last non-zero remainder \(=2\).

Answer: \(\boxed2\).

Question 7. Determine whether \(\frac{37}{2^3\times5^2}\) has a terminating decimal expansion.

1. Denominator is: \[ 2^3\times5^2 \]
2. Only prime factors \(2\) and \(5\) are present.
3. Therefore it is of the form \(2^m5^n\).

Answer: It has a terminating decimal expansion.

Question 8. Determine whether \(\frac{77}{210}\) has a terminating decimal expansion.

1. First reduce: \[ \frac{77}{210}=\frac{11}{30} \]
2. Factorise the denominator: \[ 30=2\times3\times5 \]
3. The factor \(3\) is present.
4. Therefore the denominator is not of the form \(2^m5^n\).

Answer: Non-terminating recurring decimal expansion.

Question 9. If the HCF of two numbers is \(16\), their LCM is \(240\), and one number is \(48\), find the other number.

1. Let the other number be \(x\).
2. Use: \[ \mathrm{HCF}\times\mathrm{LCM}= \text{product of numbers} \]
3. Substitute: \[ 16\times240=48x \]
4. \[ 3840=48x \]
5. Divide by \(48\): \[ x=\frac{3840}{48} \]
6. \[ x=80 \]

Answer: \(\boxed{80}\).

Question 10. Prove that \(3\sqrt2\) is irrational.

1. Assume \(3\sqrt2\) is rational.
2. Since \(3\) is a non-zero rational number, divide by \(3\).
3. Then: \[ \sqrt2=\frac{3\sqrt2}{3} \]
4. If \(3\sqrt2\) were rational, then the quotient of two rational numbers \(3\sqrt2\) and \(3\) would be rational.
5. This would make \(\sqrt2\) rational.
6. But \(\sqrt2\) is irrational.
7. This is a contradiction.

Hence \(\boxed{3\sqrt2\text{ is irrational}}\).

12. Final 96% Target Revision

Must-Remember Formulas

\[ \boxed{a=bq+r,\qquad0\le r

\[ \boxed{\mathrm{HCF}(a,b)=\mathrm{HCF}(b,r)} \]

\[ \boxed{\mathrm{HCF}(a,b)\times\mathrm{LCM}(a,b)=a\times b} \]

\[ \boxed{q=2^m5^n\Rightarrow\text{terminating decimal}} \]

\[ \boxed{\text{Other prime factor in reduced denominator}\Rightarrow\text{non-terminating recurring}} \]

Board-Exam Proof Checklist

1. For Euclid questions, write every division line clearly.
2. Stop only when the remainder becomes \(0\).
3. Take the last non-zero remainder as HCF.
4. For prime factorisation, continue until every factor is prime.
5. For HCF use minimum powers; for LCM use maximum powers.
6. For decimal questions, reduce the fraction first.
7. For irrationality proofs, start with “Assume, to the contrary…”
8. Write the contradiction clearly before the final conclusion.

Common Mistakes to Avoid

• Using the decimal denominator test before reducing the fraction.
• Taking maximum powers for HCF instead of minimum powers.
• Forgetting that the remainder must be less than the divisor.
• Calling every square root irrational, even \(\sqrt{25}=5\).
• Skipping the coprime condition in irrationality proofs.
• Stopping Euclid's algorithm before the remainder becomes zero.
• Writing only the final answer in a step-marking question.

96% Marks Target – Practise These Most

• Euclid's division lemma and algorithm.
• HCF questions using Euclid's algorithm.
• Word problems involving remainders.
• Prime factorisation.
• HCF and LCM using prime factors.
• HCF × LCM relationship.
• Proofs that \(\sqrt2,\sqrt3,\sqrt5\) are irrational.
• Proofs involving rational + irrational expressions.
• Terminating and non-terminating recurring decimal expansions.
• Mixed board-style questions requiring more than one concept.

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Class 10 Mathematics • Chapter 1: Real Numbers

Detailed handwritten-style notes for simple learning and board-exam preparation.