CBSE Class 10 Mathematics Real Numbers Assignment Set 03

Euclid's Division Lemma or Euclid's Division Algorithm

If you have any two positive integers \( a \) and \( b \), there will always exist unique whole numbers \( q \) and \( r \) that satisfy the relationship:
\( a = bq + r \), where \( 0 \le r < b \)

In this equation, \( a \) represents the dividend, \( b \) is the divisor, \( q \) is the quotient, and \( r \) is the remainder. This is commonly written as:
Dividend = (Divisor \( \times \) Quotient) + Remainder

Algorithm

An algorithm refers to a sequence of clearly defined steps that outlines the procedure for solving a specific mathematical problem.

Lemma

A lemma is an already established mathematical statement that is employed to prove a subsequent theorem or statement.

Euclid's Division Algorithm

This is a specific mathematical method used to calculate or determine the Highest Common Factor (HCF) of any two positive integers.

Fundamental Theorem of Arithmetic

This theorem states that any composite number can be uniquely represented as a product of prime numbers, disregarding the order in which these prime factors are written. For instance, the number \( 98280 \) can be broken down using prime factorization:

Consequently, we can write \( 98280 = 2^3 \times 3^3 \times 5 \times 7 \times 13 \) as a product of powers of prime numbers.

Rational Numbers and Their Decimal Expansion

i. For a rational number \( a = \frac{p}{q} \), if \( p \) and \( q \) are coprime integers, and the prime factorization of the denominator \( q \) is of the form \( 2^n \times 5^m \) (where both \( n \) and \( m \) are non-negative integers), then \( a \) will possess a terminating decimal expansion.

ii. Conversely, if the prime factorization of \( q \) cannot be expressed in the form \( 2^n \times 5^m \), then the decimal expansion of \( a \) will be non-terminating and repeating (recurring).

Level-I
 

Question. Euclid’s Division lemma states that for any two positive integers a and b, there exist unique integers q and r such that a=bq+r, where r must satisfy.
Answer: The remainder \( r \) must be greater than or equal to zero, but strictly less than the divisor \( b \), which is mathematically written as:
\( 0 \le r < b \)
In simple words: When you divide one positive number by another, the leftover remainder can be zero or any number smaller than the one you divided by.
Exam Tip: Remember that the remainder can never be equal to or greater than the divisor, nor can it be negative.

Question. Express 10010 and 140 as prime factors.
Answer: The prime factorization of both numbers is as follows:
For \( 10010 \): \( 10010 = 2 \times 5 \times 7 \times 11 \times 13 \)
For \( 140 \): \( 140 = 2^2 \times 5 \times 7 \)
In simple words: Break down each number into a product of only prime numbers.
Exam Tip: Always write your final prime factorization with prime bases in ascending order, using exponents for repeating factors like \( 2^2 \).

Question. If p/q is a rational number (q≠0), what is the condition of q so that the decimal representation of p/q is terminating?
Answer: The denominator \( q \) must be expressible in the form \( 2^n \times 5^m \), where \( n \) and \( m \) are non-negative integers.
In simple words: The bottom part of the fraction, when broken down, should only contain the prime numbers 2 or 5.
Exam Tip: Be careful to state that \( n \) and \( m \) are non-negative integers (whole numbers), as either exponent can be zero.

Question. Write any one rational number between √2 and √3.
Answer: A suitable rational number lying between \( \sqrt{2} \) (approximately \( 1.414 \)) and \( \sqrt{3} \) (approximately \( 1.732 \)) is \( 1.5321 \).
In simple words: We can find any simple fraction or terminating decimal that sits between 1.414 and 1.732.
Exam Tip: Any terminating decimal, like 1.5 or 1.6, is a simple and reliable answer to write in your exam.

Question. Find the [HCF x LCM] for 105 and 120.
Answer: Using the fundamental relationship that the product of the HCF and LCM of two numbers is equal to the product of those two numbers, we get:
\( \text{HCF}(105, 120) \times \text{LCM}(105, 120) = 105 \times 120 \)
\( \implies \text{HCF} \times \text{LCM} = 12600 \)
In simple words: Multiplying the HCF and LCM of two numbers gives the exact same result as multiplying the two numbers themselves.
Exam Tip: Do not waste time calculating the individual HCF and LCM; simply multiply the two given numbers to get the answer.

Question. If two numbers are 26 and 91 and their H.C.F is 13 then LCM is.
Answer: We can calculate the least common multiple by dividing the product of the two numbers by their HCF:
\( \text{LCM}(26, 91) = \frac{26 \times 91}{\text{HCF}(26, 91)} \)
\( \implies \text{LCM} = \frac{26 \times 91}{13} \)
\( \implies \text{LCM} = 2 \times 91 = 182 \)
In simple words: You can find the LCM by multiplying the two numbers together and then dividing by their highest common factor.
Exam Tip: Use cancellation (like dividing 26 by 13 to get 2) to make the multiplication much faster and avoid errors.

Question. The decimal expansion of the rational number 33/2^2.5 will terminate.
Answer: For the rational number \( \frac{33}{2^2 \times 5} \), the denominator has a maximum power of \( 2 \) (which is \( 2^2 \)). Therefore, the decimal expansion of this number will terminate after exactly two decimal places:
\( \frac{33}{2^2 \times 5} = \frac{33 \times 5}{2^2 \times 5^2} = \frac{165}{100} = 1.65 \)
In simple words: Since the highest power of 2 or 5 in the denominator is 2, the decimal will stop after two digits.
Exam Tip: The number of decimal places after which a rational fraction terminates is equal to the higher power between \( 2^n \) and \( 5^m \) in the denominator.

Question. HCF of two consecutive integers x and x+1 is.
Answer: Two consecutive integers, \( x \) and \( x + 1 \), do not share any common factors other than \( 1 \). Consequently, their Highest Common Factor (HCF) is always \( 1 \).
In simple words: Any two numbers that sit right next to each other, like 4 and 5, only have 1 as their common divisor.
Exam Tip: Remember that consecutive integers are always co-prime, meaning their HCF is always \( 1 \).

Level-II
 

Question. Use Euclid division algorithm to find the HCF of:- i) 196 and 38220 ii) 867 and 255
Answer:
i) Since \( 38220 > 196 \), we apply Euclid's division lemma:
\( 38220 = 196 \times 195 + 0 \)
Because the remainder is zero, the divisor is our HCF.
\( \implies \text{HCF}(196, 38220) = 196 \)

ii) Since \( 867 > 255 \), we apply Euclid's division lemma:
\( 867 = 255 \times 3 + 102 \)
Since the remainder \( 102 \ne 0 \), we apply the lemma again to \( 255 \) and \( 102 \):
\( 255 = 102 \times 2 + 51 \)
Since the remainder \( 51 \ne 0 \), we apply the lemma to \( 102 \) and \( 51 \):
\( 102 = 51 \times 2 + 0 \)
Now that the remainder is \( 0 \), the divisor at this step is the HCF.
\( \implies \text{HCF}(867, 255) = 51 \)
In simple words: We divide the larger number by the smaller one, then divide the divisor by the remainder, repeating this until the remainder becomes zero. The final divisor is the HCF.
Exam Tip: Always write down the complete steps of Euclid's division algorithm and explicitly state the remainder at each stage to secure full marks.

Question. Find the greatest common factor of 2730 and 9350.
Answer: We can find the greatest common factor (GCF) by using Euclid's division algorithm:
\( 9350 = 2730 \times 3 + 1160 \)
\( 2730 = 1160 \times 2 + 410 \)
\( 1160 = 410 \times 2 + 340 \)
\( 410 = 340 \times 1 + 70 \)
\( 340 = 70 \times 4 + 60 \)
\( 70 = 60 \times 1 + 10 \)
\( 60 = 10 \times 6 + 0 \)
Since the remainder is now zero, the greatest common factor is \( 10 \).
In simple words: The biggest number that can divide both 2730 and 9350 without leaving any remainder is 10.
Exam Tip: Greatest Common Factor (GCF) is another name for Highest Common Factor (HCF); use the same division method to solve it.

Question. IF HCF (90,144)=18 find LCM (90,144)
Answer: We use the standard formula relating HCF and LCM of two numbers:
\( \text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)} \)
Substituting the given values:
\( \text{LCM}(90, 144) = \frac{90 \times 144}{18} \)
\( \implies \text{LCM}(90, 144) = 5 \times 144 = 720 \)
In simple words: To find the LCM, multiply the two numbers and divide the result by their HCF.
Exam Tip: Always check your final LCM by verifying that it is a multiple of both original numbers.

Question. Is 7x5x3x2+3 is a composite number? Justify your answer.
Answer: Let us simplify the given expression first:
\( 7 \times 5 \times 3 \times 2 + 3 = 3 \times (7 \times 5 \times 2 + 1) \)
\( \implies 3 \times (70 + 1) = 3 \times 71 = 213 \)
Since the number \( 213 \) can be expressed as a product of two prime numbers, \( 3 \) and \( 71 \), it has factors other than \( 1 \) and itself. Therefore, it is a composite number.
In simple words: Since we can factor out a 3 from the whole sum, the final number has 3 as a divisor and is therefore composite.
Exam Tip: Whenever you need to justify why a number is composite, factor out a common term to show it has divisors other than 1.

Question. Write 98 as product of its prime factors.
Answer: We can express \( 98 \) as the product of its prime factors:
\( 98 = 2 \times 49 \)
\( \implies 98 = 2 \times 7 \times 7 \)
\( \implies 98 = 2 \times 7^2 \)
In simple words: Divide 98 by the smallest prime numbers until you cannot divide any further.
Exam Tip: Expressing the prime factors with exponents, like \( 7^2 \), is highly recommended for presentation.

Question. Find the largest number which divides 245 and 1245 leaving remainder 5 in each case.
Answer: To find the largest number that divides both \( 245 \) and \( 1245 \) with a remainder of \( 5 \) in each case, we subtract the remainder from each number first:
\( 245 - 5 = 240 \)
\( 1245 - 5 = 1240 \)
Now, we calculate the Highest Common Factor (HCF) of \( 240 \) and \( 1240 \) using Euclid's division algorithm:
\( 1240 = 240 \times 5 + 40 \)
\( 240 = 40 \times 6 + 0 \)
The HCF of \( 240 \) and \( 1240 \) is \( 40 \). Thus, the required largest number is \( 40 \).
In simple words: Subtract 5 from both numbers first, then find the highest common factor of the remaining numbers.
Exam Tip: Always subtract the respective remainders from the given numbers before finding the HCF in such problems.

Question. There is a circular path around a sports field. Geeta takes 20 minutes to drive one round of the field. While Ravi takes 14 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at staring point?
Answer: To find the time when Geeta and Ravi will meet again at the starting point, we calculate the Least Common Multiple (LCM) of their round times, \( 20 \) and \( 14 \):
First, find the prime factorization of both numbers:
\( 20 = 2^2 \times 5 \)
\( 14 = 2 \times 7 \)
The LCM is found by taking the highest power of each prime factor:
\( \text{LCM}(20, 14) = 2^2 \times 5 \times 7 \)
\( \implies \text{LCM} = 4 \times 5 \times 7 = 140 \)
Hence, they will meet again at the starting point after \( 140 \) minutes.
In simple words: We find the smallest number that can be divided by both 20 and 14, which is 140 minutes.
Exam Tip: In meeting-point problems on a circular track, the time to meet at the start is always the LCM of the individual lap times.

Level-III
 

Question. Using Euclid’s division lemma, show that the cube of any positive integer is of the form 9q, 9q+1 or 9q+8 for some integer ‘q’.
Answer: Let \( a \) be any positive integer. Using Euclid's division lemma with a divisor of \( 3 \), we can represent \( a \) in one of three forms:
\( a = 3m \), \( a = 3m + 1 \), or \( a = 3m + 2 \) (where \( m \) is an integer).

We now cube each of these cases:

Case 1: When \( a = 3m \)
\( a^3 = (3m)^3 = 27m^3 \)
\( \implies a^3 = 9(3m^3) \)
\( \implies a^3 = 9q \) (where \( q = 3m^3 \) is an integer).

Case 2: When \( a = 3m + 1 \)
\( a^3 = (3m + 1)^3 = 27m^3 + 27m^2 + 9m + 1 \)
\( \implies a^3 = 9(3m^3 + 3m^2 + m) + 1 \)
\( \implies a^3 = 9q + 1 \) (where \( q = 3m^3 + 3m^2 + m \) is an integer).

Case 3: When \( a = 3m + 2 \)
\( a^3 = (3m + 2)^3 = 27m^3 + 54m^2 + 36m + 8 \)
\( \implies a^3 = 9(3m^3 + 6m^2 + 4m) + 8 \)
\( \implies a^3 = 9q + 8 \) (where \( q = 3m^3 + 6m^2 + 4m \) is an integer).

Thus, the cube of any positive integer is always of the form \( 9q, 9q + 1, \) or \( 9q + 8 \).
In simple words: Any positive integer can be written as a multiple of 3, or with a remainder of 1 or 2. When we cube these three forms, we find they always fit into one of the three patterns: \( 9q, 9q+1, \) or \( 9q+8 \).
Exam Tip: Always use \( 3 \) as the divisor instead of \( 9 \) for this proof; it reduces the number of cases to solve from nine down to three

Question. Show that one and only one out of n,n+2,n+4 is divisible by 3.
Answer: Let \( n \) be any positive integer. By division algorithm, we can express \( n \) as \( 3k, 3k + 1, \) or \( 3k + 2 \) for some integer \( k \).

Case 1: If \( n = 3k \)
- \( n = 3k \) (divisible by 3)
- \( n + 2 = 3k + 2 \) (not divisible by 3)
- \( n + 4 = 3k + 4 = 3(k + 1) + 1 \) (not divisible by 3)
Here, only \( n \) is divisible by 3.

Case 2: If \( n = 3k + 1 \)
- \( n = 3k + 1 \) (not divisible by 3)
- \( n + 2 = 3k + 3 = 3(k + 1) \) (divisible by 3)
- \( n + 4 = 3k + 5 = 3(k + 1) + 2 \) (not divisible by 3)
Here, only \( n + 2 \) is divisible by 3.

Case 3: If \( n = 3k + 2 \)
- \( n = 3k + 2 \) (not divisible by 3)
- \( n + 2 = 3k + 4 = 3(k + 1) + 1 \) (not divisible by 3)
- \( n + 4 = 3k + 6 = 3(k + 2) \) (divisible by 3)
Here, only \( n + 4 \) is divisible by 3.

Therefore, exactly one out of \( n, n + 2, \) and \( n + 4 \) is divisible by 3 for any positive integer \( n \).
In simple words: If you take any starting number, the two numbers that are 2 and 4 steps away from it will form a trio. In this trio, exactly one number will always be in the 3 times table.
Exam Tip: Clearly outline all three possible cases for the value of \( n \) and check each expression under every case to get full marks.

Question. Prove that 5+√3 is an irrational no.
Answer: Let us assume, on the contrary, that \( 5 + \sqrt{3} \) is a rational number. Therefore, we can find coprime integers \( a \) and \( b \) (\( b \ne 0 \)) such that:
\( 5 + \sqrt{3} = \frac{a}{b} \)
Rearranging the terms, we get:
\( \sqrt{3} = \frac{a}{b} - 5 \)
\( \implies \sqrt{3} = \frac{a - 5b}{b} \)
Since \( a \) and \( b \) are integers, the expression \( \frac{a - 5b}{b} \) must be rational. This implies that \( \sqrt{3} \) is also a rational number. However, this contradicts the established fact that \( \sqrt{3} \) is irrational. Consequently, our initial assumption is incorrect, and \( 5 + \sqrt{3} \) is indeed an irrational number.
In simple words: We pretend the number is rational, which leads us to a contradiction where \( \sqrt{3} \) would also have to be rational. Since we know \( \sqrt{3} \) is irrational, our starting guess must have been wrong.
Exam Tip: In contradiction proofs, clearly state the step where the rational expression equals the irrational radical, as this is the key marking point.

Question. Using Euclid’s algorithm find the HCF of the following 4052 and 12576
Answer: We apply Euclid's division algorithm to \( 12576 \) and \( 4052 \):
\( 12576 = 4052 \times 3 + 420 \)
\( 4052 = 420 \times 9 + 272 \)
\( 420 = 272 \times 1 + 148 \)
\( 272 = 148 \times 1 + 124 \)
\( 148 = 124 \times 1 + 24 \)
\( 124 = 24 \times 5 + 4 \)
\( 24 = 4 \times 6 + 0 \)
Since the remainder is now \( 0 \), the divisor at this final stage is \( 4 \). Therefore, the HCF of \( 4052 \) and \( 12576 \) is \( 4 \).
In simple words: Keep dividing the divisor by the remainder until you get zero. The last number you divided by is the answer.
Exam Tip: Even if the division has many steps, do not skip any in your working; examiners verify every line.

Question. Show that the square of any odd integer is of the form 4q+1 for some integer q.
Answer: Let \( a \) be any odd positive integer. Any odd integer can be written in the form \( 2m + 1 \) (where \( m \) is an integer).
Squaring this expression, we get:
\( a^2 = (2m + 1)^2 \)
\( \implies a^2 = 4m^2 + 4m + 1 \)
\( \implies a^2 = 4(m^2 + m) + 1 \)
Let \( q = m^2 + m \). Since \( m \) is an integer, \( q \) must also be an integer.
\( \implies a^2 = 4q + 1 \)
Consequently, the square of any odd integer is always of the form \( 4q + 1 \).
In simple words: Every odd number can be written as one more than an even number. When we square it and factor out a 4, we are always left with a remainder of 1.
Exam Tip: Writing \( 2m+1 \) as the algebraic representation of an odd number is the standard and simplest way to start this proof.

Question. Prove that √5+√3 is irrational.
Answer: Assume, on the contrary, that \( \sqrt{5} + \sqrt{3} \) is a rational number, say \( p \).
Thus, we can write:
\( \sqrt{5} + \sqrt{3} = p \)
Squaring both sides of the equation:
\( (\sqrt{5} + \sqrt{3})^2 = p^2 \)
\( \implies 5 + 3 + 2\sqrt{15} = p^2 \)
\( \implies 8 + 2\sqrt{15} = p^2 \)
\( \implies 2\sqrt{15} = p^2 - 8 \)
\( \implies \sqrt{15} = \frac{p^2 - 8}{2} \)
Since \( p \) is a rational number, \( \frac{p^2 - 8}{2} \) must also be rational. This means \( \sqrt{15} \) is rational, which contradicts the fact that \( \sqrt{15} \) is irrational. Therefore, our assumption is false, and \( \sqrt{5} + \sqrt{3} \) is indeed an irrational number.
In simple words: If we assume the sum is rational, squaring both sides leads to the conclusion that \( \sqrt{15} \) is rational, which we know is false. Therefore, the original sum must be irrational.
Exam Tip: Squaring both sides is the most elegant way to isolate a single square root term and establish the contradiction.

Self Evaluation Questions
 

Question. Draw the factor tree for 678.
Answer: The factor tree for 678 is constructed by repeatedly dividing the composite factors by prime numbers:
\( 678 = 2 \times 339 \)
\( \implies 339 = 3 \times 113 \)
Since \( 2 \), \( 3 \), and \( 113 \) are all prime numbers, the tree is complete:
In simple words: We divide 678 by 2 to get 339, and then divide 339 by 3 to get 113, which is a prime number.
Exam Tip: In a factor tree, make sure that the leaves at the bottom are all prime numbers, showing that the number cannot be broken down any further.

Question. The sum of two numbers is 1660 and HCF is 20 find the numbers.
Answer: Let the two required numbers be \( 20x \) and \( 20y \), where \( x \) and \( y \) are coprime integers.
According to the problem, their sum is \( 1660 \):
\( 20x + 20y = 1660 \)
\( \implies 20(x + y) = 1660 \)
\( \implies x + y = 83 \)
Since \( 83 \) is a prime number, any pair of positive integers \( (x, y) \) that add up to \( 83 \) will be coprime. Thus, multiple pairs of numbers are possible. For example:
- If \( x = 1 \) and \( y = 82 \), the numbers are \( 20 \times 1 = 20 \) and \( 20 \times 82 = 1640 \).
- If \( x = 2 \) and \( y = 81 \), the numbers are \( 40 \) and \( 1620 \).
Any such pair of numbers satisfying \( 20x \) and \( 20y \) with \( x + y = 83 \) is a valid solution.
In simple words: Since the HCF is 20, both numbers must be multiples of 20. When we divide their sum by 20, we find the two remaining parts must add up to 83.
Exam Tip: Always state that the non-common factors \( x \) and \( y \) must be coprime to ensure the HCF remains exactly \( 20 \).

Question. Prove that √7 is irrational number.
Answer: Let us assume, on the contrary, that \( \sqrt{7} \) is a rational number. Therefore, we can express it as:
\( \sqrt{7} = \frac{a}{b} \)
where \( a \) and \( b \) are coprime integers and \( b \ne 0 \).
Squaring both sides:
\( 7 = \frac{a^2}{b^2} \)
\( \implies a^2 = 7b^2 \)
This means that \( 7 \) divides \( a^2 \). By the fundamental theorem, if a prime divides the square of a number, it also divides the number itself. Thus, \( 7 \) divides \( a \).
Let \( a = 7c \) for some integer \( c \). Substituting this back into our equation:
\( (7c)^2 = 7b^2 \)
\( \implies 49c^2 = 7b^2 \)
\( \implies b^2 = 7c^2 \)
This implies that \( 7 \) divides \( b^2 \), and consequently, \( 7 \) also divides \( b \).
Hence, \( a \) and \( b \) share a common factor of \( 7 \), which directly contradicts our assumption that they are coprime. Thus, \( \sqrt{7} \) is an irrational number.
In simple words: We assume the number can be written as a simplified fraction. We then prove that both the top and bottom would have to be divisible by 7, which means the fraction was not fully simplified, creating a contradiction.
Exam Tip: Ensure you mention the theorem: 'if \( p \) is a prime and divides \( a^2 \), then \( p \) divides \( a \)' to secure full marks.

Question. Prove that 3 / (2√5) is irrational number.
Answer: Let us assume that \( \frac{3}{2\sqrt{5}} \) is a rational number, which we can set equal to \( r \) (where \( r \) is rational and \( r \ne 0 \)):
\( \frac{3}{2\sqrt{5}} = r \)
Rearranging the equation to solve for \( \sqrt{5} \):
\( 2\sqrt{5} = \frac{3}{r} \)
\( \implies \sqrt{5} = \frac{3}{2r} \)
Since \( r \) is a non-zero rational number, the right-hand side \( \frac{3}{2r} \) must also be a rational number. This would mean that \( \sqrt{5} \) is rational, which is a contradiction since we know \( \sqrt{5} \) is irrational. Thus, \( \frac{3}{2\sqrt{5}} \) is an irrational number.
In simple words: If we assume the fraction is rational, we can rearrange it to show that \( \sqrt{5} \) must also be rational. Since we know \( \sqrt{5} \) is irrational, the entire fraction must be irrational.
Exam Tip: Keep the square root term isolated on one side of the equation when showing it is equal to a rational expression.

Question. Write 2/13 in decimal form and comment on decimal expansion.
Answer: Let us convert the fraction \( \frac{2}{13} \) into its decimal form by performing division:
\( \frac{2}{13} = 0.153846153846... \)
\( \implies \frac{2}{13} = 0.\overline{153846} \)
The decimal expansion has a repeating block of six digits '153846'. Therefore, we can comment that the decimal expansion of \( \frac{2}{13} \) is non-terminating and repeating.
In simple words: When we divide 2 by 13, the numbers keep going forever, but they repeat the same six-digit pattern over and over.
Exam Tip: Always use a bar over the repeating block of digits to clearly show the periodic part of the decimal.

Question. If √ab be an irrational number prove that √a + √b is irrational.
Answer: Let us assume, on the contrary, that \( \sqrt{a} + \sqrt{b} \) is a rational number, say \( x \):
\( \sqrt{a} + \sqrt{b} = x \)
Squaring both sides of this equation gives:
\( (\sqrt{a} + \sqrt{b})^2 = x^2 \)
\( \implies a + b + 2\sqrt{ab} = x^2 \)
Rearranging the equation to solve for \( \sqrt{ab} \):
\( 2\sqrt{ab} = x^2 - (a + b) \)
\( \implies \sqrt{ab} = \frac{x^2 - (a + b)}{2} \)
Since \( a \), \( b \), and \( x \) are rational numbers, the expression \( \frac{x^2 - (a + b)}{2} \) is also rational. This means \( \sqrt{ab} \) must be a rational number, which directly contradicts the given condition that \( \sqrt{ab} \) is an irrational number. Therefore, our assumption is incorrect, and \( \sqrt{a} + \sqrt{b} \) is indeed irrational.
In simple words: If we assume the sum of the square roots is rational, squaring both sides forces the term \( \sqrt{ab} \) to also be rational. Since we are told \( \sqrt{ab} \) is irrational, the sum must be irrational.
Exam Tip: Explicitly state that squaring both sides is the primary method to isolate the term \( \sqrt{ab} \) and expose the contradiction.

Question. Show that any positive odd integer will be of the form 4q+1 or 4q+3 where q is some integer.
Answer: Let \( a \) be a positive odd integer. Applying Euclid's division algorithm with a divisor of \( 4 \), we write:
\( a = 4q + r \)
where \( q \) is an integer and \( 0 \le r < 4 \). The possible values for the remainder \( r \) are \( 0, 1, 2, \) or \( 3 \). This gives us four possible forms for \( a \):
1. \( a = 4q \)
2. \( a = 4q + 1 \)
3. \( a = 4q + 2 \)
4. \( a = 4q + 3 \)

However, since \( a \) is specified to be an odd integer:
- \( 4q = 2(2q) \), which is an even integer.
- \( 4q + 2 = 2(2q + 1) \), which is also an even integer.
Since \( a \) cannot be even, it cannot be of the form \( 4q \) or \( 4q + 2 \). Therefore, any positive odd integer must be of the form \( 4q + 1 \) or \( 4q + 3 \) for some integer \( q \).
In simple words: Any whole number divided by 4 leaves a remainder of 0, 1, 2, or 3. The numbers with remainders of 0 and 2 are even, so only the ones with remainders of 1 and 3 are odd.
Exam Tip: Be sure to write down the intermediate step showing why \( 4q \) and \( 4q+2 \) are even by factoring out a \( 2 \).