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(a)(i) Explain why any integer that is a factor (or a divisor) of both \(m\) and \(n\) must also be a factor of their difference \(m-n,\) and of their sum \(m+n\). (ii) Prove that $$ H C F(m, n)=H C F(m-n, n) $$ (iii) Use this to calculate in your head \(H C F(1001,91)\) without factorising either number. (b)(i) Prove that: \(H C F(m, m+1)=1\). (ii) Find \(H C F(m, 2 m+1)\). (iii) Find \(H C F\left(m^{2}+1, m-1\right)\).

Short Answer

Expert verified
HCF(1001, 91) = 91; HCF(m, m+1) = 1; HCF(m, 2m+1) = 1; HCF(m^2+1, m-1) = 1.

Step by step solution

01

Understand the relation of factors to sums and differences

If an integer is a factor of two numbers, say \(d\) is a factor of both \(m\) and \(n\), then \(d\) divides both \(m\) and \(n\) exactly. Thus, \(m - n = (m - n) = m + (-n)\) must also be divisible by \(d\) because subtracting two multiples of \(d\) results in another multiple of \(d\). Similarly, \(m + n\) must be divisible by \(d\) because adding two multiples of \(d\) results in another multiple of \(d\).
02

Prove the Equality of HCF

To prove that \(HCF(m, n) = HCF(m-n, n)\), consider the fact that any common divisor of \(m\) and \(n\) must divide both \(m-n\) and \(n\). Conversely, any common divisor of \(m-n\) and \(n\) must divide \(m\) (since \(m = (m-n) + n\)). Thus, the set of common divisors is the same for both pairs \((m, n)\) and \((m-n, n)\), which means their HCFs are equal.
03

Calculate HCF(1001, 91) using Step 2

From Step 2, \(HCF(1001, 91) = HCF(1001-91, 91) = HCF(910, 91)\). Next, calculate \(HCF(910, 91)\). Note that 910 is a multiple of 91 (specifically \(910 = 91 \times 10\)), so \(HCF(910, 91) = 91\). Thus, \(HCF(1001, 91) = 91\).
04

Prove HCF(m, m+1) is 1

Since \(m\) and \(m+1\) are consecutive integers, they have no common factors other than 1. Therefore, their highest common factor is 1, i.e., \(HCF(m, m+1) = 1\).
05

Find HCF(m, 2m+1)

Using the same reasoning as in previous steps, since \(m\) and \(2m+1\) are not multiples of each other and have no common factors (since any common factor dividing \(m\) and \(2m+1\) must divide their difference, which would reduce to 1), we get \(HCF(m, 2m+1) = 1\).
06

Find HCF(m^2+1, m-1)

Use the identity \((m^2 + 1) - (m-1)(m+1) = 2\) to observe that \(HCF(m^2+1, m-1)\) divides 2. If \(m-1 = 1\), then \(HCF(m^2+1, m-1) = 1\). For general \(m\), small divisors can be found by examining cases but generally, if both are greater than 1, using properties and specific evaluations shows \(HCF(m^2+1, m-1) = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Greatest Common Divisor
How do you determine the largest integer that divides two or more integers without a remainder? This is what the Greatest Common Divisor (GCD), also known as the Highest Common Factor (HCF), is concerned with. The GCD of two integers, say \( a \) and \( b \), is the largest integer \( d \) that divides both \( a \) and \( b \) completely. Understanding the concept of GCD is important because it helps simplify fractions and solve a broad range of number theory problems.

To determine the GCD, you can use the Euclidean Algorithm, which involves division, subtraction, or repeated use of the remainder operations between two numbers until you find a zero remainder. The other number, at this zero remainder step, is the GCD. For example, in the solution provided, the formula \( HCF(m, n) = HCF(m-n, n) \) utilizes this idea by showing that the subtraction of two numbers does not alter their common divisors, hence keeping their GCD unchanged.

This reveals a critical insight: the GCD is invariant under certain integer operations, which are immensely useful for mental calculations as seen with \( HCF(1001, 91) \). Without factoring directly, you can apply this understanding to simplify calculations effortlessly.
Properties of Integers
Integers are whole numbers that can be positive, negative, or zero. They possess certain properties that are crucial when solving mathematical problems involving the GCD and other concepts in number theory. These properties include closure, associative, commutative, and distributive properties.

When dealing with the division of integers, it is important to note the associative and commutative properties help in reordering and grouping numbers without affecting the outcome. The solution above leverages the inherent properties of integers by showing how operations like addition and subtraction of multiples maintain divisibility.

For example, if \( d \) is a divisor of both \( m \) and \( n \), the sum \( m + n \) and difference \( m - n \) are also divisible by \( d \). This intuitive property helps us prove relationships between integers and their GCDs without diving into complex factorization processes, assisting in simpler and more efficient problem-solving strategies.
Divisibility
Divisibility refers to the ability of one integer to be divided by another without leaving a remainder. In number theory, determining divisibility helps in identifying factors or divisors of numbers, making it easier to solve problems involving the GCD.

The principle of divisibility underpins many proofs and is particularly significant in understanding how the GCD remains constant even when operations such as sum and difference are performed. Notice how in the solution, when a common divisor divides both \( m \) and \( n \), it automatically divides \( (m-n) \) and \( (m+n) \) as well. This is because when you subtract or add two multiples of a number, the result is also a multiple of that number.

This straightforward concept is powerful in evaluating divisibility quickly, allowing mathematicians and students alike to verify and comprehend number relationships easily without stress.
Mathematical Proofs
Mathematical proofs are logical arguments that demonstrate the truth of a given proposition. In mathematics, especially number theory, constructing proofs is a crucial skill, as it involves not just deriving results but explaining why they are true.

Proofs often use assumptions, definitions, and previously established results to deduce new truths. In the exercise, the proof that \( HCF(m, n) = HCF(m-n, n) \) is a classic example. It relies on the deductive step-by-step logic showing that because all common divisors of \( m \) and \( n \) must divide both \( m-n \) and \( n \), and the reverse holds true, the two pairs have the same GCD.

Developing proofs like these strengthens your mathematical reasoning skills and offers insight into deeper properties of numbers, encouraging a greater appreciation for the elegant structure within mathematical systems.

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Most popular questions from this chapter

(a) Joining the midpoints of the edges of an equilateral triangle \(A B C\) cuts the triangle into four identical smaller equilateral triangles. Removing one of the three outer small triangles (say \(A M N\), with \(M\) on \(A C\) ) leaves three-quarters of the original shape in the form of an isosceles trapezium \(M N B C .\) Show how to cut this isosceles trapezium into four congruent pieces. (b) Joining the midpoints of opposite sides of a square cuts the square into four congruent smaller squares. If we remove one of these squares, we are left with three-quarters of the original square in the form of an L-shape. Show how to cut this L-shape into four congruent pieces.

In how many different ways can the missing digits in this short multiplication be completed? $$ \begin{array}{r} \quad \square 6 \\ \times \quad \square \\ \hline \square 28 \\ \hline \end{array} $$ One would like students not only to master the direct operation of multiplying digits effectively, but also to notice that the inverse procedure of "identifying the multiples of a given integer that give rise to a specified output" depends on the HCF of the multiplier and the base (10) of the numeral system. \- Multiplying by \(1,3,7,\) or 9 induces a one-to-one mapping on the set of ten digits \(0-9\); so an inverse problem such as \(" 7 \times \square\) ends in 6 " has just one digit-solution. \- Multiplying by \(2,4,6,\) or 8 induces a two-to-one mapping onto the set of even digits (multiples of 2); so an inverse problem such as " \(6 \times \square\) ends in 4" has two digit-solutions, and an inverse problem such as " \(6 \times \square\) ends in 3 " has no digit-solutions. \- Multiplying by 5 induces a five-to-one mapping onto the multiples (0 and 5 ) of 5 , so an inverse problem such as " \(5 \times \square\) ends in 0 " has five digit-solutions and an inverse problem such as " \(5 \times \square\) ends in 3 " has no digit-solutions at all. \- Multiplying by 0 induces a ten-to-one mapping onto the multiples of 0 (namely 0 ); so an inverse problem such as " \(0 \times \square\) ends in 0 " has ten digit-solutions and an inverse problem such as " \(0 \times \square\) ends in 3 (or any digit other than 0 )" has no digit-solutions at all. The next problem shows - in a very simple setting - how elusive inverse problems can be. Here, instead of being asked to perform a direct calculation, the rules and the answer are given, and we are simply asked to invent a calculation that gives the specified output.

Imagine a triangle \(A B C\) on the unit sphere (with radius \(r=\) 1), with angle \(\alpha\) between \(A B\) and \(A C,\) angle \(\beta\) between \(B C\) and \(B A,\) and angle \(\gamma\) between \(C A\) and \(C B\). You are now in a position to derive the remarkable formula for the area of such a spherical triangle. (a) Let the two great circles containing the sides \(A B\) and \(A C\) meet again at \(A^{\prime} .\) If we imagine \(A\) as being at the North pole, then \(A^{\prime}\) will be at the South pole, and the angle between the two great circles at \(A^{\prime}\) will also be \(\alpha .\) The slice contained between these two great circles is called a lune with angle \(\alpha\) (i) What fraction of the surface area of the whole sphere is contained in this lune of angle \(\alpha ?\) Write an expression for the actual area of this lune. (ii) If the sides \(A B\) and \(A C\) are extended backwards through \(A,\) these backward extensions define another lune with the same angle \(\alpha,\) and the same surface area. Write down the total area of these two lunes with angle \(\alpha\) (b)(i) Repeat part (a) for the two sides \(B A, B C\) meeting at the vertex \(B,\) to find the total area of the two lunes meeting at \(B\) and \(B^{\prime}\) with angle \(\beta\). (ii) Do the same for the two sides \(C A, C B\) meeting at the vertex \(C,\) to find the total area of the two lunes meeting at \(C\) and \(C^{\prime}\) with angle \(\gamma\). (c)(i) Add up the areas of these six lunes (two with angle \(\alpha,\) two with angle \(\beta,\) and two with angle \(\gamma\) ). Check that this total includes every part of the sphere at least once. (ii) Which parts of the sphere have been covered more than once? How many times have you covered the area of the original triangle \(A B C ?\) And how many times have you covered the area of its sister triangle \(A^{\prime} B^{\prime} C^{\prime} ?\) (iii) Hence find a formula for the area of the triangle \(A B C\) in terms of its angles \(-\alpha\) at \(A, \beta\) at \(B,\) and \(\gamma\) at \(C\)

The three integers \(a=3, b=4, c=5\) in the Pythagorean triple (3,4,5) form an arithmetic progression: that is, \(c-b=b-a\). Find all Pythagorean triples \((a, b, c)\) which form an arithmetic progression \(-\) that is, for which \(c-b=b-a\)

Let \(A B C\) be a triangle. We use the standard labelling convention, whereby the side \(B C\) opposite \(A\) has length \(a\), the side \(C A\) opposite \(B\) has length \(b,\) and the side \(A B\) opposite \(C\) has length \(c\). Prove that, if \(c^{2}=a^{2}+b^{2},\) then \(\angle B C A\) is a right angle.

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