Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 1: Problem 2

Translate the following expression into prefix form $$ \frac{5+4+\left(2-\left(3-\left(6+\frac{4}{5}\right)\right)\right)}{3(6-2)(2-7)} $$

Short Answer

Expert verified
`/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))) (* 3 (- 6 2) (- 2 7)))`

Step by step solution

01

Understand the Expression

The given expression is \( \frac{5+4+(2-(3-(6+\frac{4}{5})))}{3(6-2)(2-7)} \). It includes a combination of addition, subtraction, fractions, and multiplication.
02

Simplify the Denominator

First, simplify the denominator: \( 3(6-2)(2-7) = 3 \cdot 4 \cdot (-5) = -60 \).
03

Simplify the Numerator

Next, simplify the numerator: Start with the innermost operation, \( 6 + \frac{4}{5} = \frac{30}{5} + \frac{4}{5} = \frac{34}{5} \). Then, \( 3 - \frac{34}{5} = \frac{15}{5} - \frac{34}{5} = -\frac{19}{5} \). Continue with \( 2 - \left(-\frac{19}{5}\right) = 2 + \frac{19}{5} = \frac{10}{5} + \frac{19}{5} = \frac{29}{5} \). Finally, \( 5 + 4 + \frac{29}{5} = 9 + \frac{29}{5} = \frac{45}{5} + \frac{29}{5} = \frac{74}{5} \).
04

Write in Prefix Notation

Combine the simplified forms in prefix notation. The operation to perform first is the division of the numerator by the denominator: \( \frac{74}{5} \div (-60) \). In prefix notation, it becomes: `/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))) (* 3 (- 6 2) (- 2 7)))`.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Expression Simplification
Expression simplification is a key step for reducing a complex mathematical expression into a more manageable form. In our exercise, we first handled the denominator and numerator separately.

Why simplify? Simplification helps in understanding the structure of an expression. It allows us to perform operations on simpler terms which makes further steps more straightforward.

In the exercise, we started by simplifying the denominator, which involved straightforward multiplication:
  • Multiply the constants: \( 3, 4, (-5) \)
  • This gives \( 3 \times 4 = 12 \), then \( 12 \times (-5) = -60 \)
This simplification allows us to focus on the numerator next. Breaking down the numerator involved nested operations:

  • Resolve fractions inside parentheses: \( 6 + \frac{4}{5} \)
  • Handle subtraction and addition as indicated by parentheses
  • Combine like terms to reach \( \frac{74}{5} \)
After simplification, each element of the problem becomes easier to work with in subsequent steps.
Arithmetic Operations
Arithmetic operations are fundamental for solving mathematical expressions. They include addition, subtraction, multiplication, and division. For our exercise, proper sequencing of operations was crucial.

Basic operations such as these work according to precedence and associativity rules:
  • Multiplication and division should be resolved before addition and subtraction.
  • Operations in parentheses are prioritized, signaling steps needed within a calculation.
In our solution:

  • We first addressed parenthetical operations, like calculating \( 6 + \frac{4}{5} \), before moving outward.
  • Arithmetic on fractions involved common denominators for operations like \( 3 - \frac{34}{5} \).
  • Finally, combining simplified results led to cleanly executing the final division of the whole expression.
Mastering these operations is essential, as they form the backbone of many mathematical problem-solving techniques.
Mathematical Notation
Mathematical notation is the language of mathematics, and it involves symbolic representations such as numbers, operations, and punctuation. In this exercise, we used a specific notational form called prefix notation.

Prefix notation, also known as Polish notation, places operators before operands. This has a particular advantage: it removes the need for parentheses to indicate order of operations, thus reducing potential for errors.

Normal expression, like our exercise:
\[\frac{5+4+(2-(3-(6+\frac{4}{5})))}{3(6-2)(2-7)} \]
Translates into a prefix notation:
  • Start with the division operator, as it encompasses the overall operation.
  • Describe each part sequentially: `/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))) (* 3 (- 6 2) (- 2 7)))`.
Understanding mathematical notation is critical for clear and effective communication of mathematical ideas. It not only aids in problem solving but also in sharing solution approaches across different domains.

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

The good-enough? test used in computing square roots will not be very effective for finding the square roots of very small numbers. Also, in real computers, arithmetic operations are almost always performed with limited precision. This makes our test inadequate for very large numbers. Explain these statements, with examples showing how the test fails for small and large numbers. An alternative strategy for implementing good-enough? is to watch how guess changes from one iteration to the next and to stop when the change is a very small fraction of the guess. Design a square-root procedure that uses this kind of end test. Does this work better for small and large numbers? \({ }^{24}\) Readers who are worried about the efficiency issues involved in using procedure calls to implement iteration should note the remarks on "tail recursion" in section 1.2.1.

Louis Reasoner is having great difficulty doing exercise \(1.24\). His fast- prime? test seems to run more slowly than his prime? test. Louis calls his friend Eva Lu Ator over to help. When they examine Louis's code, they find that he has rewritten the expmod procedure to use an explicit multiplication, rather than calling square: (define (expmod base exp m) (cond \(((=\exp 0) 1)\) \(((\) even? exp) \((\) remainder \((*(\) expmod base \((/ \exp 2) \mathrm{m})\) \((\) expmod base \((/ \exp 2) \mathrm{m}))\) \((\) else \(\quad\) m)) \((\) remainder \((*\) base \((\) expmod base \((-\exp 1) \mathrm{m}))\) \(\quad\) m) \())\) "I don't see what difference that could make," says Louis. "I do." says Eva. "By writing the procedure like that, you have transformed the \(\Theta(\log n)\) process into a \(\Theta(n)\) process." Explain.

Newton's method for cube roots is based on the fact that if \(y\) is an approximation to the cube root of \(x\), then a better approximation is given by the value \(\frac{x / y^{2}+2 y}{3}\) Use this formula to implement a cube-root procedure analogous to the squareroot procedure. (In section \(1.3 .4\) we will see how to implement Newton's method in general as an abstraction of these square-root and cube- root procedures.)

A function \(f\) is defined by the rule that \(f(n)=n\) if \(n<3\) and \(f(n)=\) \(f(n-1)+2 f(n-2)+3 f(n-3)\) if \(n \geq 3\). Write a procedure that computes \(f\) by means of a recursive process. Write a procedure that computes \(f\) by means of an iterative process.

Ben Bitdiddle has invented a test to determine whether the interpreter he is faced with is using applicative-order evaluation or normal-order evaluation. He defines the following two procedures: (define \((\mathrm{p})(\mathrm{p}))\) (define (test \(\mathrm{x} \mathrm{y}\) ) (if \((=\mathrm{x} 0)\) 0 \(\mathrm{y})\) ) Then he evaluates the expression (test \(0(\mathrm{p}))\) What behavior will Ben observe with an interpreter that uses applicative-order evaluation? What behavior will he observe with an interpreter that uses normalorder evaluation? Explain your answer. (Assume that the evaluation rule for the special form if is the same whether the interpreter is using normal or applicative order: The predicate expression is evaluated first, and the result determines whether to evaluate the consequent or the alternative expression.)
See all solutions

Recommended explanations on Computer Science Textbooks

Fintech

Read Explanation

Issues in Computer Science

Read Explanation

Functional Programming

Read Explanation

Computer Programming

Read Explanation

Blockchain Technology

Read Explanation

Databases

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.

Sign-up for free