1. Euclid’s Division Lemma

1. Use Euclid's Division Lemma to find the HCF of:

   • (i) 135 and 225
   • (ii) 196 and 38220
   • (iii) 867 and 255

Euclid’s Division Lemma

Euclid’s Division Lemma states that for any two positive integers aa and bb (a>ba > b), there exist unique integers qq (quotient) and rr (remainder) such that:

a=bq+rwhere0≤r

The process is repeated using bb and rr until r=0r = 0. At that point, the divisor (bb) becomes the HCF of aa and bb.

Solution

(i) HCF of 135 and 225

Let a=225a = 225 and b=135b = 135. Applying Euclid’s Division Lemma:

1. Step 1: Divide 225 by 135

   225=135⋅1+90225 = 135 \cdot 1 + 90

   Here, the remainder r=90r = 90. Replace aa with 135 and bb with 90.

2. Step 2: Divide 135 by 90

   135=90⋅1+45135 = 90 \cdot 1 + 45

   Here, r=45r = 45. Replace aa with 90 and bb with 45.

3. Step 3: Divide 90 by 45

   90=45⋅2+090 = 45 \cdot 2 + 0

   Here, r=0r = 0. The process stops, and b=45b = 45 is the HCF.

HCF of 135 and 225 = 45

(ii) HCF of 196 and 38220

Let a=38220a = 38220 and b=196b = 196. Applying Euclid’s Division Lemma:

1. Step 1: Divide 38220 by 196 38220=196⋅195+038220 = 196 \cdot 195 + 0 Here, r=0r = 0. The process stops immediately, and b=196b = 196 is the HCF.

HCF of 196 and 38220 = 196

(iii) HCF of 867 and 255

Let a=867a = 867 and b=255b = 255. Applying Euclid’s Division Lemma:

1. Step 1: Divide 867 by 255

   867=255⋅3+102867 = 255 \cdot 3 + 102

   Here, r=102r = 102. Replace aa with 255 and bb with 102.

2. Step 2: Divide 255 by 102

   255=102⋅2+51255 = 102 \cdot 2 + 51

   Here, r=51r = 51. Replace aa with 102 and bb with 51.

3. Step 3: Divide 102 by 51

   102=51⋅2+0102 = 51 \cdot 2 + 0

   Here, r=0r = 0. The process stops, and b=51b = 51 is the HCF.

HCF of 867 and 255 = 51

Key Takeaways

1. The Euclid’s Division Lemma involves repeated division and finding remainders until the remainder is 0.
2. The last non-zero divisor is the HCF of the two numbers.

Final Results

1. HCF of 135 and 225 = 45
2. HCF of 196 and 38220 = 196
3. HCF of 867 and 255 = 51

This systematic approach ensures the accuracy of results while highlighting the simplicity of Euclid's Division Lemma.

2. Prove that the square of any positive integer is of the form 3m3m or 3m+13m + 1, for some integer mm.

Explanation

We need to show that the square of any positive integer can be written in the form 3m3m or 3m+13m + 1. To do this, we use the division algorithm to express a positive integer nn in terms of its remainder when divided by 3.

Step 1: Represent any integer nn using division by 3

For any positive integer nn, when divided by 3, there are three possible remainders: 0, 1, or 2. This means any integer nn can be written as:

n=3q,n=3q+1,orn=3q+2n = 3q, \quad n = 3q + 1, \quad \text{or} \quad n = 3q + 2

where qq is an integer.

Step 2: Square each case

We now square nn in each of the three cases:

1. Case 1: n=3qn = 3q
   Squaring nn, we get: n2=(3q)2=9q2=3(3q2)n^2 = (3q)^2 = 9q^2 = 3(3q^2) This is clearly divisible by 3, so n2n^2 is of the form 3m3m, where m=3q2m = 3q^2.

2. Case 2: n=3q+1n = 3q + 1
   Squaring nn, we get: n2=(3q+1)2=9q2+6q+1=3(3q2+2q)+1n^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 Here, n2n^2 is of the form 3m+13m + 1, where m=3q2+2qm = 3q^2 + 2q.

3. Case 3: n=3q+2n = 3q + 2
   Squaring nn, we get: n2=(3q+2)2=9q2+12q+4=3(3q2+4q+1)+1n^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 Again, n2n^2 is of the form 3m+13m + 1, where m=3q2+4q+1m = 3q^2 + 4q + 1.

Step 3: Conclusion

From the above cases, we see that:

1. When n=3qn = 3q, n2n^2 is of the form 3m3m.
2. When n=3q+1n = 3q + 1 or n=3q+2n = 3q + 2, n2n^2 is of the form 3m+13m + 1.

Thus, the square of any positive integer nn is either of the form 3m3m or 3m+13m + 1.

Key Takeaways

• Using the division algorithm, any integer can be expressed in terms of remainders when divided by 3.
• Squaring each possible case proves that the square of any positive integer is either divisible by 3 or leaves a remainder of 1 when divided by 3.

This completes the proof.

2. HCF and LCM

3. Find the HCF and LCM of 12 and 18 using the prime factorization method. Verify that $\text{HCF}\times \text{LCM}=\text{Product of the numbers}$.

Solution :- 

Step 1: Prime Factorization

We express each number as a product of its prime factors:

• 12=2×2×3=22×3112 = 2 \times 2 \times 3 = 2^2 \times 3^1
• 18=2×3×3=21×3218 = 2 \times 3 \times 3 = 2^1 \times 3^2

Step 2: Find the HCF (Highest Common Factor)

The HCF is obtained by multiplying the smallest power of all common prime factors.

From the prime factorizations:

• The common prime factors are 22 and 33.
• The smallest powers are 212^1 (from 1212 and 1818) and 313^1 (from 1212 and 1818).

Thus:

HCF=21×31=2×3=6\text{HCF} = 2^1 \times 3^1 = 2 \times 3 = 6

Step 3: Find the LCM (Least Common Multiple)

The LCM is obtained by multiplying the highest power of all prime factors.

From the prime factorizations:

• The highest powers are 222^2 (from 1212) and 323^2 (from 1818).

Thus:

LCM=22×32=4×9=36\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = 36

Step 4: Verify the Relationship

The relationship between HCF, LCM, and the product of the two numbers is:

HCF×LCM=Product of the numbers\text{HCF} \times \text{LCM} = \text{Product of the numbers}

Here:

• HCF=6\text{HCF} = 6
• LCM=36\text{LCM} = 36
• Product of the numbers=12×18=216\text{Product of the numbers} = 12 \times 18 = 216

Now, calculate:

HCF×LCM=6×36=216\text{HCF} \times \text{LCM} = 6 \times 36 = 216

Since 216=216216 = 216, the relationship is verified.

Final Results

1. HCF of 12 and 18 = 6
2. LCM of 12 and 18 = 36
3. Verification: HCF ×\times LCM = Product of the numbers

Thus, the solution is complete, and the relationship holds true.

4. Two tankers contain 720 liters and 405 liters of diesel. Find the maximum capacity of a container that can measure both quantities exactly.

Solution :- 

To find the maximum capacity of a container that can measure both 720 liters and 405 liters of diesel exactly, we need to determine the HCF (Highest Common Factor) of 720 and 405. The HCF represents the largest quantity that can divide both 720 and 405 exactly.

Step 1: Prime Factorization

Let's first find the prime factorization of both numbers.

1. Prime factorization of 720:

720÷2=360720 \div 2 = 360 360÷2=180360 \div 2 = 180 180÷2=90180 \div 2 = 90 90÷2=4590 \div 2 = 45 45÷3=1545 \div 3 = 15 15÷3=515 \div 3 = 5 5÷5=15 \div 5 = 1

So, the prime factorization of 720 is:

720=24×32×5720 = 2^4 \times 3^2 \times 5

2. Prime factorization of 405:

405÷3=135405 \div 3 = 135 135÷3=45135 \div 3 = 45 45÷3=1545 \div 3 = 15 15÷3=515 \div 3 = 5 5÷5=15 \div 5 = 1

So, the prime factorization of 405 is:

405=34×5405 = 3^4 \times 5

Step 2: Find the HCF

Now, we find the HCF by taking the lowest power of each common prime factor.

• The common prime factors are 33 and 55.
• For 33, the lowest power is 323^2 (from 720).
• For 55, the lowest power is 515^1 (common in both 720 and 405).

So, the HCF is:

HCF=32×5=9×5=45\text{HCF} = 3^2 \times 5 = 9 \times 5 = 45

Step 3: Conclusion

The maximum capacity of a container that can measure both 720 liters and 405 liters exactly is 45 liters.

3. Fundamental Theorem of Arithmetic

5. Express each of the following numbers as a product of their prime factors:

   • (i) 140
   • (ii) 156
   • (iii) 3825

Solution :- 

The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime number itself or can be expressed as a product of prime numbers, and this factorization is unique, except for the order of the factors.

We will use this theorem to express each of the given numbers as a product of their prime factors.

(i) Prime Factorization of 140

Start by dividing 140 by the smallest prime number, 2, and continue factoring:

140÷2=70140 \div 2 = 70 70÷2=3570 \div 2 = 35 35÷5=735 \div 5 = 7 7÷7=17 \div 7 = 1

So, the prime factorization of 140 is:

140=22×5×7140 = 2^2 \times 5 \times 7

(ii) Prime Factorization of 156

Start by dividing 156 by the smallest prime number, 2:

156÷2=78156 \div 2 = 78 78÷2=3978 \div 2 = 39 39÷3=1339 \div 3 = 13 13÷13=113 \div 13 = 1

So, the prime factorization of 156 is:

156=22×3×13156 = 2^2 \times 3 \times 13

(iii) Prime Factorization of 3825

Start by dividing 3825 by the smallest prime number, 3, and continue factoring:

3825÷3=12753825 \div 3 = 1275 1275÷3=4251275 \div 3 = 425 425÷5=85425 \div 5 = 85 85÷5=1785 \div 5 = 17 17÷17=117 \div 17 = 1

So, the prime factorization of 3825 is:

3825=32×52×173825 = 3^2 \times 5^2 \times 17

Final Results

• Prime factorization of 140: 22×5×72^2 \times 5 \times 7
• Prime factorization of 156: 22×3×132^2 \times 3 \times 13
• Prime factorization of 3825: 32×52×173^2 \times 5^2 \times 17

These are the prime factorizations of the given numbers.

6. Check whether 6n^nn can end with the digit 0 for any natural number n.

Solution :- 

We need to check whether 6n6^n can end with the digit 0 for any natural number nn.

Understanding the Problem

For a number to end with the digit 0, it must be divisible by 10. Hence, we need to check if 6n6^n can be divisible by 10 for any natural number nn.

Step 1: Divisibility by 10

A number is divisible by 10 if it is divisible by both 2 and 5. Therefore, for 6n6^n to end in 0, it must be divisible by both 2 and 5.

• Divisibility by 2: Since 6 is even, 6n6^n is always divisible by 2 for any n≥1n \geq 1.
• Divisibility by 5: However, 6 is not divisible by 5, and raising 6 to any power will still not make it divisible by 5. Specifically, 6n6^n will never include a factor of 5.

Step 2: Conclusion

Since 6n6^n is never divisible by 5, it can never be divisible by 10. Therefore, 6n6^n cannot end with the digit 0 for any natural number nn.

Thus, the answer is no, 6n6^n can never end with the digit 0 for any natural number nn.

4. Rational and Irrational Numbers

7. Show that any positive odd integer is of the form $4q+1$ or $4q+3$, where q is some integer.

Solution:

We need to prove that any positive odd integer can be expressed in one of the following two forms:

(i)4q+1or(ii)4q+3\text{(i)} \quad 4q + 1 \quad \text{or} \quad \text{(ii)} \quad 4q + 3

where qq is some integer.

Step 1: Understanding Odd Integers

An odd integer is any integer that is not divisible by 2. It can be written in the general form:

odd integer=2k+1for some integer k.\text{odd integer} = 2k + 1 \quad \text{for some integer} \, k.

This represents any odd number (such as 1, 3, 5, 7, etc.).

Step 2: Divide the Odd Integer by 4

Now, we take any odd integer and divide it by 4. Since every integer can be expressed in the form 4q+r4q + r, where qq is the quotient and rr is the remainder, we will consider the possible remainders when dividing an odd number by 4.

• When dividing an integer by 4, the remainder can only be one of the following values: 0,1,2,or30, 1, 2, \text{or} 3.
• However, since we are considering odd integers, the remainder cannot be 0 or 2 because these would indicate even numbers. Therefore, the remainder must be either 1 or 3.

Thus, the odd integer can be expressed as:

odd integer=4q+1or4q+3\text{odd integer} = 4q + 1 \quad \text{or} \quad 4q + 3

where qq is some integer (the quotient from the division).

Step 3: Conclusion

Since any positive odd integer when divided by 4 results in a remainder of either 1 or 3, we have shown that any positive odd integer is of the form 4q+14q + 1 or 4q+34q + 3, where qq is some integer.

Thus, the proof is complete.

8. Prove that 2\sqrt{2} is an irrational number.

Solution:

To prove that 2\sqrt{2} is irrational, we will use proof by contradiction. This means we will assume the opposite of what we want to prove, and show that it leads to a contradiction.

Step 1: Assume 2\sqrt{2} is rational

Suppose, for the sake of contradiction, that 2\sqrt{2} is a rational number. By definition of rational numbers, if 2\sqrt{2} is rational, then it can be expressed as the ratio of two integers aa and bb, where aa and bb are coprime (i.e., they have no common factors other than 1), and b≠0b \neq 0.

Thus, we can write:

2=ab\sqrt{2} = \frac{a}{b}

where aa and bb are integers, and gcd⁡(a,b)=1\gcd(a, b) = 1.

Step 2: Square both sides of the equation

Next, we square both sides of the equation:

22=(ab)2\sqrt{2}^2 = \left( \frac{a}{b} \right)^2 2=a2b22 = \frac{a^2}{b^2}

Now, multiply both sides by b2b^2 to eliminate the denominator:

2b2=a22b^2 = a^2

This means that a2a^2 is even, because it is two times some integer b2b^2.

Step 3: Conclusion about aa

Since a2a^2 is even, it follows that aa must also be even (because the square of an odd number is odd). Therefore, aa is divisible by 2, and we can write:

a=2ka = 2k

for some integer kk.

Step 4: Substitute a=2ka = 2k into the equation

Substitute a=2ka = 2k into the equation 2b2=a22b^2 = a^2 from Step 2:

2b2=(2k)22b^2 = (2k)^2 2b2=4k22b^2 = 4k^2

Now, divide both sides of the equation by 2:

b2=2k2b^2 = 2k^2

This shows that b2b^2 is also even, and therefore bb must be even as well (because, similarly, the square of an odd number is odd).

Step 5: Contradiction

At this point, we have shown that both aa and bb are even, which means both are divisible by 2. But this contradicts our assumption that aa and bb are coprime (i.e., that gcd⁡(a,b)=1\gcd(a, b) = 1).

Since we reached a contradiction, our assumption that 2\sqrt{2} is rational must be false.

Step 6: Conclusion

Since assuming that 2\sqrt{2} is rational leads to a contradiction, we conclude that 2\sqrt{2} is irrational.

Thus, the proof is complete.

9. Prove that 3+253 + 2\sqrt{5} is irrational.

Solution:

We will prove by contradiction that 3+253 + 2\sqrt{5} is irrational.

Step 1: Assume 3+253 + 2\sqrt{5} is rational

Suppose, for the sake of contradiction, that 3+253 + 2\sqrt{5} is rational. By definition of rational numbers, a rational number can be expressed as the ratio of two integers pp and qq, where pp and qq are coprime (i.e., they have no common factors other than 1) and q≠0q \neq 0.

Thus, we assume:

3+25=pq3 + 2\sqrt{5} = \frac{p}{q}

where pp and qq are integers, and gcd⁡(p,q)=1\gcd(p, q) = 1.

Step 2: Isolate 252\sqrt{5}

Next, subtract 3 from both sides of the equation to isolate 252\sqrt{5}:

25=pq−32\sqrt{5} = \frac{p}{q} - 3

Express 3 as a fraction:

25=pq−3qq=p−3qq2\sqrt{5} = \frac{p}{q} - \frac{3q}{q} = \frac{p - 3q}{q}

Step 3: Simplify and solve for 5\sqrt{5}

Now, divide both sides of the equation by 2:

5=p−3q2q\sqrt{5} = \frac{p - 3q}{2q}

At this point, we have expressed 5\sqrt{5} as a ratio of two integers p−3qp - 3q and 2q2q. This means that 5\sqrt{5} is rational because it is written as a ratio of integers.

Step 4: Contradiction

However, we know that 5\sqrt{5} is an irrational number (it cannot be expressed as a ratio of two integers, as shown in the proof for 2\sqrt{2}). This contradicts our assumption that 5\sqrt{5} is rational.

Step 5: Conclusion

Since our assumption that 3+253 + 2\sqrt{5} is rational leads to a contradiction, we conclude that 3+253 + 2\sqrt{5} is irrational.

Thus, the proof is complete.

5. Miscellaneous Questions

10. Without performing the actual division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

    • (i) $13/3125$
    • (ii) $29/343$
    • (iii) $23/8$

$Solution:-$

To determine whether a rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion, we use the following rule:

• A rational number will have a terminating decimal expansion if its denominator (in the simplified form) has only 2 and/or 5 as prime factors.
• If the denominator contains any prime factor other than 2 or 5, the decimal expansion will be non-terminating repeating.

(i) 133125\frac{13}{3125}

First, we find the prime factorization of 3125:

$3125=553125\; =\; 5^5$

Since the denominator has only 5 as a prime factor, the decimal expansion of 133125\frac{13}{3125} will be terminating.

(ii) 29343\frac{29}{343}

Next, we find the prime factorization of 343:

$343=73343\; =\; 7^3$

Since the denominator has 7 as a prime factor (and not 2 or 5), the decimal expansion of 29343\frac{29}{343} will be non-terminating repeating.

(iii) 238\frac{23}{8}

Finally, we find the prime factorization of 8:

$8=238\; =\; 2^3$

Since the denominator has only 2 as a prime factor, the decimal expansion of 238\frac{23}{8} will be terminating.

Conclusion:

• 133125\frac{13}{3125} has a terminating decimal expansion.
• 29343\frac{29}{343} has a non-terminating repeating decimal expansion.
• 238\frac{23}{8} has a terminating decimal expansion.

11. Find the smallest number that is divisible by both 36 and 84.

Solution:- 

To find the smallest number that is divisible by both 36 and 84, we need to find their Least Common Multiple (LCM).

Step 1: Prime Factorization of 36 and 84

Start by finding the prime factorization of both numbers.

• Prime factorization of 36:

36=22×3236 = 2^2 \times 3^2

• Prime factorization of 84:

84=22×3×784 = 2^2 \times 3 \times 7

Step 2: Find the LCM

The LCM is found by taking the highest power of each prime factor that appears in the factorization of both numbers.

• For the prime factor 2, the highest power is 222^2.
• For the prime factor 3, the highest power is 323^2.
• For the prime factor 7, the highest power is 717^1.

Thus, the LCM is:

LCM=22×32×7=4×9×7=252LCM = 2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 252

Step 3: Conclusion

The smallest number that is divisible by both 36 and 84 is 252.

11 . Prove that $\sqrt{7}$ is an irrational number.

Solution :- 

Let  assume that $\sqrt{7}$ is rational. Then, there exist co-prime positive integers $a$ and $b$ such that