Real Numbers — Notes

Introduction

This chapter delves deeper into real numbers, building upon the concepts of irrational numbers introduced in Class IX. We will explore fundamental properties of positive integers, focusing on two key theorems: Euclid's division algorithm and the Fundamental Theorem of Arithmetic.

Euclid's Division Algorithm

Euclid's division algorithm is a foundational concept in number theory concerning the divisibility of integers.

• Statement: For any two positive integers $a$ (dividend) and $b$ (divisor), there exist unique integers $q$ (quotient) and $r$ (remainder) such that:

where $0\le r<b$.

• How it works: This algorithm is essentially a formalization of the standard long division process. When you divide a number $a$ by a number $b$, you get a quotient $q$ and a remainder $r$. The crucial part is that the remainder $r$ is always non-negative and strictly less than the divisor $b$.

• Why it matters: While seemingly simple, this algorithm has significant applications in understanding the properties of integers. Its primary use in this chapter will be to compute the Highest Common Factor (HCF) of two positive integers.

Example: Applying Euclid's Division Algorithm

Given: Two positive integers, 135 and 225.

To find: The HCF of 135 and 225 using Euclid's division algorithm.

Solution:

1. Since 225 > 135, we apply the division lemma to $a=225$ and $b=135$.

The remainder is 90, which is not 0.

2. Now, we take the divisor (135) as the new dividend and the remainder (90) as the new divisor.

The remainder is 45, which is not 0.

3. Again, we take the divisor (90) as the new dividend and the remainder (45) as the new divisor.

The remainder is 0.

The last non-zero remainder is 45.

Answer: The HCF of 135 and 225 is 45.

The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic deals with the multiplication of positive integers and their prime factorizations.

• Statement: Every composite number can be expressed as a product of prime numbers, and this factorization is unique, apart from the order in which the prime factors appear.

• How it works: A composite number is a whole number that has more than two factors (including 1 and itself). A prime number is a whole number greater than 1 that has only two factors: 1 and itself. The theorem states that any composite number can be broken down into a unique set of prime numbers that multiply together to give the original composite number.

• Why it matters: This theorem is a cornerstone of number theory with profound implications:

1. Proving Irrationality: It is used to prove that certain numbers, like $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$, are irrational.
2. Decimal Expansions of Rationals: It helps us determine when the decimal expansion of a rational number is terminating and when it is non-terminating repeating. This depends on the prime factorization of the denominator of the rational number.

Example: Prime Factorization

Given: The composite number 12.

To find: The unique prime factorization of 12.

Solution:

1. We start by dividing 12 by the smallest prime number, 2.

2. Now, we factorize 6. The smallest prime number that divides 6 is 2.

3. The result is 3, which is a prime number. So, the prime factors of 12 are 2, 2, and 3.

Answer: The unique prime factorization of 12 is $2\times 2\times 3$, or $2^2\times 3$.

Example: Proving Irrationality (Concept Introduction)

While the formal proofs of irrationality are covered in a later section, the Fundamental Theorem of Arithmetic provides the basis. For instance, to prove $\sqrt{2}$ is irrational, we assume it's rational ($\frac{p}{q}$), square both sides, and use prime factorization properties to show a contradiction. The theorem guarantees that the prime factorization of any integer is unique, and this uniqueness is what leads to the contradiction.

Example: Decimal Expansion of Rational Numbers

Given: The rational number $\frac{13}{8}$.

To find: The nature of its decimal expansion.

Solution:

1. We examine the prime factorization of the denominator, 8. $8=2\times 4=2\times 2\times 2=2^3$.

2. According to the Fundamental Theorem of Arithmetic, if the denominator $q$ of a rational number $\frac{p}{q}$ (in its simplest form) has only prime factors of the form $2^n\times 5^m$ (where $n$ and $m$ are non-negative integers), then its decimal expansion will be terminating.

3. Here, the denominator is $8=2^3$. This can be written as $2^3\times 5^0$. The only prime factor is 2.

Answer: Therefore, the decimal expansion of $\frac{13}{8}$ is terminating. ($\frac{13}{8}=1.625$)

Remember

The uniqueness of prime factorization is the core idea behind the Fundamental Theorem of Arithmetic. Even if you find factors in a different order, you will always end up with the same set of prime factors.

Exam-smart

Questions might involve finding the HCF of large numbers using Euclid's algorithm, or determining if a given rational number will have a terminating or non-terminating repeating decimal expansion based on its denominator's prime factors. You might be asked to explain why a number is irrational, referencing the unique prime factorization.

The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic is a cornerstone of number theory, stating that every composite number can be uniquely expressed as a product of prime numbers, disregarding the order of the factors.

A composite number is a natural number greater than 1 that has more than two distinct positive divisors (including 1 and itself). For example, 4, 6, 8, 9, 10 are composite numbers. A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. For example, 2, 3, 5, 7, 11 are prime numbers.

The theorem can be formally stated as:

Theorem 1.1 (Fundamental Theorem of Arithmetic): Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

This means that if you take any composite number, there is only one set of prime numbers that multiply together to give that composite number. For example, the prime factorization of 12 is always $2\times 2\times 3$ (or $2^2\times 3$), no matter how you arrive at it. You cannot find a different set of prime numbers that also multiply to 12.

Applications of the Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic has several important applications, particularly in finding the Highest Common Factor (HCF) and Least Common Multiple (LCM) of numbers.

Prime Factorisation Method for HCF and LCM

This method relies on expressing each number as a product of its prime factors.

• HCF: The HCF of a set of numbers is the product of the lowest powers of all common prime factors in their prime factorisations.
• LCM: The LCM of a set of numbers is the product of the highest powers of all prime factors that appear in any of the factorisations.

Example 1: Finding HCF and LCM using Prime Factorisation

Given: Two numbers, 6 and 20. To find: Their HCF and LCM.

Solution:

1. Prime Factorisation:

• $6=2\times 3=2^1\times 3^1$
• $20=2\times 2\times 5=2^2\times 5^1$

2. Identify Common Prime Factors and their Powers:

• The only common prime factor is 2.
• The lowest power of 2 is $2^1$.
• The highest power of 2 is $2^2$.
• The prime factor 3 appears only in the factorisation of 6. Its highest power is $3^1$.
• The prime factor 5 appears only in the factorisation of 20. Its highest power is $5^1$.

3. Calculate HCF:

• HCF(6, 20) = Product of lowest powers of common prime factors.
• HCF(6, 20) = $2^1=2$.

4. Calculate LCM:

• LCM(6, 20) = Product of highest powers of all prime factors involved.
• LCM(6, 20) = $2^2\times 3^1\times 5^1=4\times 3\times 5=60$.

Answer: HCF(6, 20) = 2, LCM(6, 20) = 60.

Relationship between HCF, LCM, and the Numbers

For any two positive integers $a$ and $b$, the following relationship holds:

This relationship is very useful for finding the LCM if the HCF is known, or vice versa.

Example 2: Using the HCF-LCM Relationship

Given: HCF(96, 404) = 4. To find: LCM(96, 404).

Solution:

1. Apply the formula:

• $\text{LCM}(a,b)=\frac{a\times b}{\text{HCF}(a,b)}$
• $\text{LCM}(96,404)=\frac{96\times 404}{4}$

2. Calculate:

• $\text{LCM}(96,404)=\frac{38784}{4}=9696$.

Answer: LCM(96, 404) = 9696.

Note: This relationship ($\text{HCF}(a,b)\times \text{LCM}(a,b)=a\times b$) is only true for two numbers. It does not generally hold for three or more numbers.

Example 3: HCF and LCM of Three Numbers

Given: Numbers 6, 72, and 120. To find: HCF and LCM.

Solution:

1. Prime Factorisation:

• $6=2\times 3=2^1\times 3^1$
• $72=8\times 9=2^3\times 3^2$
• $120=12\times 10=(2^2\times 3)\times (2\times 5)=2^3\times 3^1\times 5^1$

2. Calculate HCF:

• Common prime factors are 2 and 3.
• Lowest power of 2 is $2^1$.
• Lowest power of 3 is $3^1$.
• HCF(6, 72, 120) = $2^1\times 3^1=2\times 3=6$.

3. Calculate LCM:

• All prime factors involved are 2, 3, and 5.
• Highest power of 2 is $2^3$.
• Highest power of 3 is $3^2$.
• Highest power of 5 is $5^1$.
• LCM(6, 72, 120) = $2^3\times 3^2\times 5^1=8\times 9\times 5=360$.

Answer: HCF(6, 72, 120) = 6, LCM(6, 72, 120) = 360.

Example 4: Numbers ending with zero

Given: A number of the form $4^n$, where $n$ is a natural number. To find: Whether $4^n$ can end with the digit zero for any value of $n$.

Solution:

1. Condition for ending with zero: For a number to end with the digit zero, its prime factorisation must include both 2 and 5. This is because a number ending in zero is a multiple of 10, and $10=2\times 5$.

2. Prime factorisation of $4^n$:

• $4^n=(2^2{)}^n=2^{2n}$.
• The only prime factor in the factorisation of $4^n$ is 2. It does not contain the prime factor 5.

3. Conclusion: Since the prime factorisation of $4^n$ never contains the prime factor 5, $4^n$ can never be a multiple of 10. Therefore, $4^n$ can never end with the digit zero for any natural number $n$. This conclusion is guaranteed by the uniqueness part of the Fundamental Theorem of Arithmetic.

Example 5: Proving a Number is Composite

Given: The number $7\times 11\times 13+13$. To prove: This number is composite.

Solution:

1. Factor out common terms:

• The number is $7\times 11\times 13+13$.
• We can see that 13 is a common factor in both terms.
• Factor out 13: $13\times (7\times 11+1)$.

2. Simplify the expression in the parenthesis:

• $7\times 11+1=77+1=78$.

3. Rewrite the number:

• The number is $13\times 78$.

4. Check for compositeness:

• Since the number can be expressed as a product of two integers ($13$ and $78$), and both are greater than 1, the number is composite.
• (Further, $78=2\times 3\times 13$, so $7\times 11\times 13+13=13\times (2\times 3\times 13)=2\times 3\times {13}^2$).

Answer: The number $7\times 11\times 13+13$ is composite because it can be written as $13\times 78$.

Example 6: Proving Another Number is Composite

Given: The number $7\times 6\times 5\times 4\times 3\times 2\times 1+5$. To prove: This number is composite.

Solution:

1. Factor out common terms:

• The number is $7\times 6\times 5\times 4\times 3\times 2\times 1+5$.
• We can see that 5 is a common factor in both terms.
• Factor out 5: $5\times (7\times 6\times 4\times 3\times 2\times 1+1)$.

2. Simplify the expression in the parenthesis:

• $7\times 6\times 4\times 3\times 2\times 1=7\times 144=1008$.
• So, the expression in the parenthesis is $1008+1=1009$.

3. Rewrite the number:

• The number is $5\times 1009$.

4. Check for compositeness:

• Since the number can be expressed as a product of two integers ($5$ and $1009$), and both are greater than 1, the number is composite. (We also need to check if 1009 is prime. It is, but that's not necessary to prove the original number composite).

Answer: The number $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ is composite because it can be written as $5\times 1009$.

Example 7: Meeting Point on a Circular Path

Given: Sonia takes 18 minutes for one round of a circular path. Ravi takes 12 minutes for the same. They start at the same point and time, going in the same direction. To find: After how many minutes will they meet again at the starting point.

Solution:

1. Understanding the problem: Sonia completes a round every 18 minutes, so she will be at the starting point at 18, 36, 54, 72, ... minutes. Ravi completes a round every 12 minutes, so he will be at the starting point at 12, 24, 36, 48, 60, 72, ... minutes. They will meet again at the starting point when the time elapsed is a common multiple of their individual times. The first time they meet again will be at the Least Common Multiple (LCM) of their times.

2. Find the LCM of 18 and 12:

• Prime factorisation of 18: $18=2\times 9=2^1\times 3^2$.
• Prime factorisation of 12: $12=2\times 6=2^2\times 3^1$.

3. Calculate LCM:

• Highest power of 2 is $2^2$.
• Highest power of 3 is $3^2$.
• LCM(18, 12) = $2^2\times 3^2=4\times 9=36$.

Answer: They will meet again at the starting point after 36 minutes.

Remember

The uniqueness of prime factorisation is what makes the Fundamental Theorem of Arithmetic so powerful. It ensures that for any composite number, there's one and only one prime "recipe" to create it.

Exam-smart

Questions involving numbers ending in 0, 5, or 6 often test your understanding of prime factors. For example, $6^n$ will always end in 6 for any natural number $n$ because its only prime factor is 2 and 3, and $2^n$ cycles through 2, 4, 8, 6 and $3^n$ cycles through 3, 9, 7, 1. When you multiply these, the units digit will always result in a 6. A question might ask to prove $6^n$ can't end in 0 by highlighting the absence of the prime factor 5.

Try this: Find the HCF and LCM of 15, 24, and 36 using the prime factorisation method. Answer: HCF(15, 24, 36) = 3, LCM(15, 24, 36) = 360

Revisiting Irrational Numbers

In Class IX, we learned that irrational numbers cannot be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q\mathrm{\ne }0$. Together with rational numbers, they form the set of real numbers. This section focuses on proving the irrationality of certain numbers, specifically $\sqrt{2}$, $\sqrt{3}$, and combinations involving rational and irrational numbers. The proofs often employ a technique called proof by contradiction.

Proof that a prime $p$ divides $a$ if $p$ divides $a^2$

This theorem is a crucial stepping stone for proving the irrationality of square roots of primes. It states that if a prime number $p$ is a factor of the square of an integer $a$, then $p$ must also be a factor of the integer $a$ itself.

Theorem 1.2: Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.

Proof:

1. Let the prime factorization of $a$ be $a=p_1{p}_2\dots p_n$, where $p_1,p_2,\dots ,p_n$ are prime factors.
2. Then, $a^2=(p_1{p}_2\dots p_n)(p_1{p}_2\dots p_n)=p_1^2{p}_2^2\dots p_n^2$.
3. We are given that $p$ divides $a^2$.
4. According to the Fundamental Theorem of Arithmetic, $p$ must be one of the prime factors in the unique prime factorization of $a^2$.
5. The prime factors of $a^2$ are $p_1,p_2,\dots ,p_n$.
6. Therefore, $p$ must be one of $p_1,p_2,\dots ,p_n$.
7. Since $a=p_1{p}_2\dots p_n$, if $p$ is one of these primes, then $p$ must divide $a$.

Proof that $\sqrt{2}$ is irrational

This proof uses the method of proof by contradiction. We assume the opposite of what we want to prove and show that it leads to a logical inconsistency.

Theorem 1.3: $\sqrt{2}$ is irrational.

Proof:

1. Assume, for the sake of contradiction, that $\sqrt{2}$ is rational.
2. This means $\sqrt{2}$ can be written as $\frac{a}{b}$, where $a$ and $b$ are integers, $b\mathrm{\ne }0$, and $a$ and $b$ are coprime (have no common factor other than 1).
3. So, $\sqrt{2}=\frac{a}{b}$.
4. Squaring both sides gives $2=\frac{a^2}{b^2}$, which rearranges to $2b^2=a^2$.
5. This implies that $a^2$ is an even number (since it is 2 times an integer $b^2$).
6. From Theorem 1.2, if $a^2$ is divisible by the prime 2, then $a$ must also be divisible by 2.
7. So, we can write $a=2c$ for some integer $c$.
8. Substitute this into the equation $2b^2=a^2$: $2b^2=(2c{)}^2=4c^2$.
9. Dividing by 2 gives $b^2=2c^2$.
10. This implies that $b^2$ is an even number.
11. Applying Theorem 1.2 again, if $b^2$ is divisible by 2, then $b$ must also be divisible by 2.
12. We have now shown that both $a$ and $b$ are divisible by 2. This means they have a common factor of 2.
13. This contradicts our initial assumption that $a$ and $b$ are coprime.
14. Therefore, our original assumption that $\sqrt{2}$ is rational must be false.
15. Hence, $\sqrt{2}$ is irrational.

Example 1: Showing $\sqrt{2}$ is irrational.

Given: The statement that $\sqrt{2}$ is irrational.

To find: A step-by-step proof of its irrationality.

Solution:

1. Assume $\sqrt{2}$ is rational, so $\sqrt{2}=\frac{a}{b}$ for coprime integers $a,b$ with $b\mathrm{\ne }0$.
2. $\Rightarrow 2=\frac{a^2}{b^2}$
3. $\Rightarrow 2b^2=a^2$
4. This shows $a^2$ is even, so $a$ is even.
5. Let $a=2c$ for some integer $c$.
6. Substitute $a=2c$ into $2b^2=a^2$: $2b^2=(2c{)}^2$
7. $\Rightarrow 2b^2=4c^2$
8. $\Rightarrow b^2=2c^2$
9. This shows $b^2$ is even, so $b$ is even.
10. Both $a$ and $b$ are even, contradicting they are coprime.
11. Therefore, $\sqrt{2}$ is irrational.

Answer: $\sqrt{2}$ is irrational.

Proof that $\sqrt{3}$ is irrational

Similar to the proof for $\sqrt{2}$, we use proof by contradiction and rely on a generalized version of Theorem 1.2 for the prime number 3.

Theorem: $\sqrt{3}$ is irrational.

Proof:

1. Assume, for contradiction, that $\sqrt{3}$ is rational.
2. Then $\sqrt{3}=\frac{a}{b}$ for coprime integers $a,b$ with $b\mathrm{\ne }0$.
3. Squaring both sides: $3=\frac{a^2}{b^2}$, which gives $3b^2=a^2$.
4. This means $a^2$ is divisible by 3.
5. If a prime $p$ divides $a^2$, then $p$ divides $a$. So, if 3 divides $a^2$, then 3 divides $a$.
6. Thus, we can write $a=3c$ for some integer $c$.
7. Substitute $a=3c$ into $3b^2=a^2$: $3b^2=(3c{)}^2=9c^2$.
8. Dividing by 3 gives $b^2=3c^2$.
9. This means $b^2$ is divisible by 3.
10. Applying the same logic (if 3 divides $b^2$, then 3 divides $b$), $b$ is also divisible by 3.
11. We have shown that both $a$ and $b$ are divisible by 3, which contradicts our assumption that $a$ and $b$ are coprime.
12. Therefore, $\sqrt{3}$ is irrational.

Example 2: Prove that 3 is irrational.

Given: The statement that 3 is irrational.

To find: A proof of its irrationality.

Solution:

1. Assume $\sqrt{3}$ is rational, so $\sqrt{3}=\frac{a}{b}$ for coprime integers $a,b$ with $b\mathrm{\ne }0$.
2. Squaring both sides: $3=\frac{a^2}{b^2}$
3. Rearranging: $3b^2=a^2$.
4. This implies $a^2$ is divisible by 3.
5. By Theorem 1.2 (extended for $p=3$), if $a^2$ is divisible by 3, then $a$ is divisible by 3.
6. So, $a=3c$ for some integer $c$.
7. Substitute $a=3c$ into $3b^2=a^2$: $3b^2=(3c{)}^2$.
8. $\Rightarrow 3b^2=9c^2$.
9. Dividing by 3: $b^2=3c^2$.
10. This implies $b^2$ is divisible by 3.
11. Therefore, $b$ is divisible by 3.
12. Since both $a$ and $b$ are divisible by 3, they have a common factor of 3.
13. This contradicts the assumption that $a$ and $b$ are coprime.
14. Hence, $\sqrt{3}$ is irrational.

Answer: $\sqrt{3}$ is irrational.

Proof that $5-\sqrt{3}$ is irrational

This proof demonstrates a general principle: the sum or difference of a rational number and an irrational number is irrational.

Theorem: $5-\sqrt{3}$ is irrational.

Proof:

1. Assume, for contradiction, that $5-\sqrt{3}$ is rational.
2. Let $5-\sqrt{3}=r$, where $r$ is a rational number.
3. Rearrange the equation to isolate $\sqrt{3}$: $5-r=\sqrt{3}$.
4. Since 5 is rational and $r$ is assumed to be rational, their difference ($5-r$) must also be a rational number.
5. So, we have $\sqrt{3}=(\text{a rational number})$.
6. This implies that $\sqrt{3}$ is rational.
7. However, we know (from the previous proof) that $\sqrt{3}$ is irrational.
8. This is a contradiction.
9. Therefore, our initial assumption that $5-\sqrt{3}$ is rational must be false.
10. Hence, $5-\sqrt{3}$ is irrational.

Example 3: Show that $5-\sqrt{3}$ is irrational.

Given: The expression $5-\sqrt{3}$.

To find: A proof that $5-\sqrt{3}$ is irrational.

Solution:

1. Assume, for contradiction, that $5-\sqrt{3}$ is rational.
2. Let $5-\sqrt{3}=r$, where $r$ is a rational number.
3. Rearrange to isolate $\sqrt{3}$: $5-r=\sqrt{3}$.
4. Since 5 is rational and $r$ is rational, $(5-r)$ is rational.
5. This implies $\sqrt{3}$ is rational.
6. But we know $\sqrt{3}$ is irrational.
7. This is a contradiction.
8. Therefore, $5-\sqrt{3}$ is irrational.

Answer: $5-\sqrt{3}$ is irrational.

Proof that $3\sqrt{2}$ is irrational

This proof illustrates another general rule: the product of a non-zero rational number and an irrational number is irrational.

Theorem: $3\sqrt{2}$ is irrational.

Proof:

1. Assume, for contradiction, that $3\sqrt{2}$ is rational.
2. Let $3\sqrt{2}=r$, where $r$ is a rational number.
3. Since $3$ is a non-zero rational number, we can divide by 3: $\sqrt{2}=\frac{r}{3}$.
4. Since $r$ is rational and $3$ is a non-zero rational number, their quotient $\frac{r}{3}$ must also be a rational number.
5. So, we have $\sqrt{2}=(\text{a rational number})$.
6. This implies that $\sqrt{2}$ is rational.
7. However, we know that $\sqrt{2}$ is irrational.
8. This is a contradiction.
9. Therefore, our initial assumption that $3\sqrt{2}$ is rational must be false.
10. Hence, $3\sqrt{2}$ is irrational.

Example 4: Show that $3\sqrt{2}$ is irrational.

Given: The expression $3\sqrt{2}$.

To find: A proof that $3\sqrt{2}$ is irrational.

Solution:

1. Assume, for contradiction, that $3\sqrt{2}$ is rational.
2. Let $3\sqrt{2}=r$, where $r$ is a rational number.
3. Since $3$ is a non-zero rational number, we can divide by 3: $\sqrt{2}=\frac{r}{3}$.
4. Since $r$ is rational and $3$ is rational, $\frac{r}{3}$ is rational.
5. This implies $\sqrt{2}$ is rational.
6. But we know $\sqrt{2}$ is irrational.
7. This is a contradiction.
8. Therefore, $3\sqrt{2}$ is irrational.

Answer: $3\sqrt{2}$ is irrational.

Remember

The proofs for irrationality rely on assuming the number is rational and then showing this leads to a contradiction with a known fact (e.g., the number is coprime, or a known irrational number is rational).

Exam-smart

For "show that" questions, clearly state your assumption and the final contradiction. You will be given an expression and asked to prove its irrationality. The steps will mirror the examples shown. You might also be given a proof and asked to identify the contradiction.

Try this: Prove that $7\sqrt{5}$ is irrational. Answer: Assume $7\sqrt{5}$ is rational, so $7\sqrt{5}=\frac{a}{b}$. Then $\sqrt{5}=\frac{a}{7b}$, which implies $\sqrt{5}$ is rational. This contradicts that $\sqrt{5}$ is irrational. Thus, $7\sqrt{5}$ is irrational.