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[This problem requires a mixture of serious thought and written proof.] (a) I choose six integers between 10 and 19 (inclusive). (i) Prove that some pair of integers among my chosen six must be relatively prime. (ii) Is it also true that some pair must have a common factor? (b) I choose six integers in the nineties (from \(90-99\) inclusive). (i) Prove that some pair among my chosen integers must be relatively prime. (ii) Is it also true that some pair must have a common factor? (c) I choose \(n+1\) integers from a run of \(2 n\) consecutive integers. (i) Prove that some pair among the chosen integers must be relatively prime. (ii) Is it also true that some pair must have a common factor?

Short Answer

Expert verified
(a)(i) Yes, relatively prime; (ii) Yes, common factor. (b)(i) Yes, relatively prime; (ii) Yes, common factor. (c)(i) Yes, relatively prime; (ii) Yes, common factor.

Step by step solution

01

Understand the Problem

We have three different subproblems involving choosing a set of integers within a specific range and analyzing their properties, focusing on relatively prime pairs and those with common factors.
02

Focus on Problem (a)(i)

Select six integers between 10 and 19. These numbers are between 10 and 19 inclusive, giving us 10 possible numbers: 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. To prove that some pair of integers among my chosen six must be relatively prime, notice that directly consecutive pairs, such as (11,12), are often relatively prime. Notably, (11, 12) is indeed relatively prime because 11 is prime, and 12 is not a multiple of 11.
03

Problem (a)(ii) Analysis

We need to prove that some pair has a common factor. Take any even number like 10 and any other even number like 12; both are divisible by 2, hence sharing a common factor.
04

Problem (b)(i) for Numbers in the Nineties

Select six integers from 90 to 99. For any six numbers within this range, consider the set of numbers: 91, 92, 93, 94, 95, 96, 97, 98, 99. Since 97 is a prime number, picking it along with any non-multiple of 97 assures a relatively prime pair.
05

Problem (b)(ii) for Common Factor in the Nineties

Many numbers are even (e.g., 92, 94, 96, 98), and any pair of even numbers will have 2 as a common factor. So, yes, some pair has a common factor.
06

Problem (c)(i) for Consecutive Integers

Consider choosing n+1 integers from a sequence of 2n consecutive integers, say from 1 to 2n. Consecutive numbers, such as (1, 2) or any similar sequence, ensure that at least one pair like (x,x+1) will be relatively prime.
07

Problem (c)(ii) for Common Factor in Consecutive Integers

In 2n consecutive numbers, some could be multiples of a particular number (e.g., two even numbers), so you can find pairs that share a common factor.

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Key Concepts

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

Relatively Prime Integers
Relatively prime integers are two numbers that share no common factors other than 1. This means they do not have any other positive integer besides 1 that can divide both of them without leaving a remainder.
A useful way to find relatively prime pairs is by checking consecutive integers, where one number is often a prime while the other is a non-multiple of the former.
For example:
  • The numbers 11 and 12 are relatively prime because 11 is a prime number, and 12 cannot be divided by 11 without a remainder.
  • In general, if you pick enough numbers from any sequence, like six numbers from 10 to 19 or the nineties, you are likely to find a relatively prime pair by concentrating on one prime number and another which is not a multiple.
Understanding relatively prime integers helps in solving number theory problems where determining the absence of common factors is crucial.
Common Factors
Common factors occur when two numbers have a positive integer greater than 1 that divides both numbers completely. Recognizing common factors is important in problems where you need to identify shared divisors.
For instance:
  • When numbers are chosen randomly, such as 10 and 12, you find they both share the common factor 2, since both numbers are divisible by 2.
  • In ranges like the nineties, pairs like 92 and 94 are even, hence, both are divisible by 2.
This indicates that in any odd-numbered set of integers, focusing on even numbers is a straightforward way to find common factors. So, whenever dealing with sets of integers, especially consecutive ones, look for simplest shared divisors to find common factors easily.
Consecutive Integers
Consecutive integers are numbers that follow each other in order. These are like 10, 11, 12,... and so on. They play a crucial role in number theory because they automatically satisfy certain properties.
Consecutive integers include both odd and even numbers, creating natural situations where some numbers are bound to be relatively prime, especially when one is an odd or a prime number.
For example:
  • Choosing any number, say 15, and the next one, 16, you'll find them to be relatively prime as 15 does not share a divisor with 16 other than 1.
  • If you choose numbers like 2 and 3, they are again relatively prime, and you find such pairs repeatedly across any run of consecutive integers.
Hence, when dealing with consecutive integers, you'll often find relatively prime pairs by default, yet also can easily identify common factors, especially when even numbers are involved.

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Most popular questions from this chapter

The 4 by 4 "multiplication table" below is completely familiar. \(\begin{array}{rrrr}1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \\ 4 & 8 & 12 & 16\end{array}\) What is the total of all the numbers in the 4 by 4 square? How should one write this answer in a way that makes the total obvious?

The "DIN A" series of paper sizes is determined by two conditions. The basic requirement is that all the DIN A rectangles are similar; the second condition is that when we fold a given size exactly in half, we get the next smaller size. Hence \- a sheet of paper of size \(\mathrm{A} 3\) folds in half to give a sheet of size \(\mathrm{A} 4-\) which is similar to \(\mathrm{A} 3 ;\) and \- a sheet of size \(A 4\) folds in half to give a sheet of size \(A 5 ;\) etc.. (a) Find the constant ratio \(r=\) "(longer side length) : (shorter side length)" for all DIN A paper sizes. (b)(i) To enlarge A4 size to A3 size (e.g. on a photocopier), each length is enlarged by a factor of \(r\). What is the "enlargement factor" to get from A3 size back to A4 size? (ii) To "enlarge" A4 size to A5 size (e.g. on a photocopier), each length is "enlarged" by a factor of \(\frac{1}{r} .\) What is the enlargement factor to get from A5 size back to A4 size?

(a) Factorise 12345 as a product of primes. (b) Using only mental arithmetic, make a list of all prime numbers up to 100 . (c)(i) Find a prime number which is one less than a square. (ii) Find another such prime. There are 4 prime numbers less than \(10 ; 25\) prime numbers less than \(100 ;\) and 168 prime numbers less than 1000 . Problem \(\mathbf{4}(\mathrm{c})\) is included to emphasise a frequently neglected message: Words and images are part of the way we communicate. But most of us cannot calculate with words and images. To make use of mathematics, we must routinely translate words into symbols. For example, unknown numbers need to be represented by symbols, and points in a geometric diagram need to be properly labelled, before we can begin to calculate, and to reason, effectively.

(a) Expand and simplify in your head: (i) \((\sqrt{2}+1)^{2}\) (ii) \((\sqrt{2}-1)^{2}\) (iii) \((1+\sqrt{2})^{3}\) (b) Simplify: (i) \(\sqrt{10+4 \sqrt{6}}\) (ii) \(\sqrt{5+2 \sqrt{6}}\) (iii) \(\sqrt{\frac{3+\sqrt{5}}{2}}\) (iv) \(\sqrt{10-2 \sqrt{5}}\) The expressions which occur in exercises to develop fluency in working with surds often appear arbitrary. But they may not be. The arithmetic of surds arises naturally: for example, some of the expressions in the previous problem have already featured in Problem \(\mathbf{3}(\mathrm{c}) .\) In particular, surds will feature whenever Pythagoras' Theorem is used to calculate lengths in geometry, or when a proportion arising from similar triangles requires us to solve a quadratic equation. So surd arithmetic is important. For example: \- A regular octagon with side length 1 can be surrounded by a square of side \(\sqrt{2}+1\) (which is also the diameter of its incircle); so the area of the regular octagon equals \((\sqrt{2}+1)^{2}-1\) (the square minus the four corners). \- \(\sqrt{2}-1\) features repeatedly in the attempt to apply the Euclidean algorithm, or anthyphairesis, to express \(\sqrt{2}\) as a "continued fraction". -\(\sqrt{10-2 \sqrt{5}}\) may look like an arbitrary, uninteresting repeated surd, but is in fact very interesting, and has already featured as \(4 \sin 36^{\circ}\) in Problem \(\mathbf{3}(\mathrm{c})\) \- One of the simplest ruler and compasses constructions for a regular pentagon \(A B C D E\) (see Problem 185) starts with a circle of radius 2 , centre \(O,\) and a point \(A\) on the circle, and in three steps constructs the next point \(B\) on the circle, where \(\underline{A B}\) is an edge of the inscribed regular pentagon, and $$ \underline{A B}=\sqrt{10-2 \sqrt{5}} $$

(a) The operation of "squaring" is a function: it takes a single real number \(x\) as input, and delivers a definite real number \(x^{2}\) as output. \- Every positive number arises as an output ("is the square of something" ). \(-\) Since \(x^{2}=(-x)^{2},\) each output (other than 0 ) arises from at least two different inputs. \- If \(a^{2}=b^{2},\) then \(0=a^{2}-b^{2}=(a-b)(a+b)\), so either \(a=b\), or \(a=-b\). Hence no two positive inputs have the same square, so each output (other than 0 ) arises from exactly two inputs (one positive and one negative). \- Hence each positive output \(y\) corresponds to just one positive input, called \(\sqrt{y}\). Find: (i) \(\sqrt{49}\) (ii) \(\sqrt{144}\) (iii) \(\sqrt{441}\) (iv) \(\sqrt{169}\) (v) \(\sqrt{196}\) (vi) \(\sqrt{961}\) (vii) \(\sqrt{96100}\) (b) Let \(a>0\) and \(b>0\). Then \(\sqrt{a b}>0\), and \(\sqrt{a} \times \sqrt{b}>0\), so both expressions are positive. Moreover, they have the same square, since $$ (\sqrt{a b})^{2}=a b=(\sqrt{a})^{2} \cdot(\sqrt{b})^{2}=(\sqrt{a} \times \sqrt{b})^{2} $$ \(\therefore \sqrt{a \times b}=\sqrt{a} \times \sqrt{b}\) Use this fact to simplify the following: (i) \(\sqrt{8}\) (ii) \(\sqrt{12}\) (iii) \(\sqrt{50}\) (iv) \(\sqrt{147}\) (v) \(\sqrt{288}\) (vi) \(\sqrt{882}\) (c) [This part requires some written calculation.] Exact expressions involving square roots occur in many parts of elementary mathematics. We focus here on just one example - namely the regular pentagon. Suppose that a regular pentagon \(A B C D E\) has sides of length \(1 .\) (i) Prove that the diagonal \(A C\) is parallel to the side \(E D\). (ii) If \(A C\) and \(B D\) meet at \(X,\) explain why \(A X D E\) is a rhombus. (iii) Prove that triangles \(A D X\) and \(C B X\) are similar. (iv) If \(A C\) has length \(x\), set up an equation and find the exact value of \(x\). (v) Find the exact length of \(B X\). (vi) Prove that triangles \(A B D\) and \(B X A\) are similar. (vii) Find the exact values of \(\cos 36^{\circ}, \cos 72^{\circ}\). (viii) Find the exact values of \(\sin 36^{\circ}, \sin 72^{\circ}\).

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