/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 (a) Using only mental arithmetic... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 11

(a) Using only mental arithmetic: (i) Determine which is bigger: $$ \frac{1}{2}+\frac{1}{5} \quad \text { or } \quad \frac{1}{3}+\frac{1}{4} ? $$ (ii) How is this question related to the observation that \(10<12 ?\) (b) [This part will require some written calculation and analysis.] (i) For positive real numbers \(x\), compare $$ \frac{1}{x+2}+\frac{1}{x+5} \quad \text { and } \quad \frac{1}{x+3}+\frac{1}{x+4} $$ (ii) What happens in part (i) if \(x\) is negative?

Short Answer

Expert verified
(a) \( \frac{1}{2} + \frac{1}{5} \) is larger; related by \( 10<12 \). (b) \( \frac{1}{x+2} + \frac{1}{x+5} > \frac{1}{x+3} + \frac{1}{x+4} \) if \( x > 0 \); differs if \( x < 0 \).

Step by step solution

01

Compare fractions for part (a)(i)

First, calculate \( \frac{1}{2} + \frac{1}{5} \). Find a common denominator, which is 10: \( \frac{1}{2} = \frac{5}{10} \) and \( \frac{1}{5} = \frac{2}{10} \), so \( \frac{1}{2} + \frac{1}{5} = \frac{5}{10} + \frac{2}{10} = \frac{7}{10} \). Next, calculate \( \frac{1}{3} + \frac{1}{4} \). The common denominator is 12: \( \frac{1}{3} = \frac{4}{12} \) and \( \frac{1}{4} = \frac{3}{12} \), so \( \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \). Compare \( \frac{7}{10} \) and \( \frac{7}{12} \). Since 10 is less than 12, \( \frac{7}{10} > \frac{7}{12} \). Therefore, \( \frac{1}{2} + \frac{1}{5} \) is larger.
02

Relate to the observation for part (a)(ii)

The observation \( 10 < 12 \) is related to the comparison of the fractions \( \frac{7}{10} \) and \( \frac{7}{12} \) because comparably smaller denominators lead to larger fractional values when numerators are equal. Hence, \( \frac{7}{10} > \frac{7}{12} \).
03

Find expression for part (b)(i)

To compare \( \frac{1}{x+2} + \frac{1}{x+5} \) and \( \frac{1}{x+3} + \frac{1}{x+4} \), note that both terms of \( \frac{1}{x+2} \) and \( \frac{1}{x+5} \) have smaller denominators than the corresponding terms of \( \frac{1}{x+3} \) and \( \frac{1}{x+4} \). This makes each individual term larger, thus making the entire sum \( \frac{1}{x+2} + \frac{1}{x+5} \) larger than \( \frac{1}{x+3} + \frac{1}{x+4} \) for positive \( x \).
04

Evaluate if x is negative for part (b)(ii)

If \( x \) is negative, then values such as \( x+2 \), \( x+5 \), \( x+3 \), or \( x+4 \) could alternate in being negative or positive, leading to a range of possibilities that affect the comparison of fractions' values. Since division by negative numbers flips inequalities, comparison would depend on the specific value of \( x \), thus significantly altering the initial assessment for positive \( x \). Specific analysis is required per negative \( x \) value.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Mental Arithmetic
Mental arithmetic is a powerful skill that enables us to solve math problems quickly in our minds without writing them down or using a calculator. When comparing fractions through mental arithmetic, it's essential to focus on estimating and simplifying the fractions to understand their relative sizes. For example, to compare \( \frac{1}{2} + \frac{1}{5} \) with \( \frac{1}{3} + \frac{1}{4} \), consider the approximate value of each sum.
  • \( \frac{1}{2} + \frac{1}{5} \) is approximately \( 0.5 + 0.2 = 0.7 \)
  • \( \frac{1}{3} + \frac{1}{4} \) is roughly \( 0.33 + 0.25 = 0.58 \)
This rough estimation suggests \( \frac{1}{2} + \frac{1}{5} \) is larger, which can be verified through traditional calculation. By practicing mental arithmetic, you get better at quickly judging the size of fractions relative to each other.
Common Denominators
Finding a common denominator is an essential technique when comparing fractions or performing addition on them. A common denominator allows you to easily compare or add fractions by giving them the same 'denominator base.' Once the fractions share a denominator, you operate directly on the numerators. For example:
  • To add \( \frac{1}{2} \) and \( \frac{1}{5} \), the common denominator is 10: \( \frac{5}{10} + \frac{2}{10} = \frac{7}{10} \).
  • For \( \frac{1}{3} \) and \( \frac{1}{4} \), the common denominator is 12: \( \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \).
Using common denominators clarifies which fraction is larger and makes the calculations more straightforward and intuitive, especially when dealing with multiple fractions in operations.
Inequalities in Fractions
Inequalities in fractions help us understand which of two fractional quantities is greater. The process involves comparing their values directly once they have a common denominator. If two fractions have the same numerator, the fraction with the smaller denominator is the larger fraction. This is because dividing by a smaller number results in a larger quotient.
For example, once \( \frac{1}{2} + \frac{1}{5} = \frac{7}{10} \) and \( \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \) are computed, compare \( \frac{7}{10} \) and \( \frac{7}{12} \). Since 10 is less than 12, \( \frac{7}{10} \) is greater than \( \frac{7}{12} \). Thus, knowing how to set up and solve these inequalities is crucial in fraction comparison problems.
Positive and Negative Numbers in Fractions
Positive and negative numbers affect the value of fractions significantly. With positive numbers in denominations, the general rule is the smaller the denominator, the larger the value of the fraction, provided numerators are equal. However, if the numbers involved become negative, the approach changes.
When comparing fractions such as \( \frac{1}{x+2} \) and \( \frac{1}{x+5} \) when \( x \) is negative, the denominators themselves may also become negative, creating more complex scenarios. Conversely, when fractions have negative values, the fraction with the larger negative denominator results in a smaller absolute fraction value because dividing by a larger (in magnitude) negative number gives a less negative (thus larger) result.
This concept requires careful attention to the sign of each term, and evaluating each case individually ensures accurate comparison.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Using only mental arithmetic: (a) Compute for yourself, and learn by heart, the times tables up to \(9 \times 9\). (b) Calculate instantly: (i) \(0.004 \times 0.02\) (ii) \(0.0008 \times 0.07\) (iii) \(0.007 \times 0.12\) (iv) \(1.08 \div 1.2\) (v) \((0.08)^{2}\) Multiplication tables are important for many reasons. They allow us to appreciate directly, at first hand, the efficiency of our miraculous place value system \(-\) in which representing any number, and implementing any operation, are reduced to a combined mastery of (i) the arithmetical behaviour of the ten digits \(0-9\), and (ii) the index laws for powers of 10 . Fluency in mental and written arithmetic then leaves the mind free to notice, and to appreciate, the deeper patterns and structures which may be lurking just beneath the surface.

[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?

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?

(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 $$ \underline{A B}=\sqrt{10-2 \sqrt{5}} $$

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 \(C\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.