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Chapter 3: Problem 3

Give a reason for your answer in each of 1-13. Assume that all variables represent integers. Does \(5 \mid 0\) ?

Short Answer

Expert verified
Yes, \(5 \mid 0\) because any number multiplied by 0 is equal to 0, and thus 0 can be expressed as a multiple of 5 without any remainder: \(0 = 5 \times 0\).

Step by step solution

01

Understand the division symbol

The symbol \(5 \mid 0\) stands for "5 divides 0." In this case, we need to find if 5 properly divides 0 without leaving any remainder.
02

Consider properties of zero

Recall that if any divisor is multiplied by zero, the result will always be zero. This implies that zero can be expressed as a multiple of any number: \[ x \times 0 = 0 \]
03

Confirm whether 5 divides 0

To check that 5 divides 0, it must hold that there exists an integer k, for which 0 = 5k. In this case, by letting the integer k be 0: \[ 0 = 5 \times 0 \] As we can see, we can indeed express 0 as the result of 5 times an integer (k = 0). Therefore, 5 does divide 0, and we can write \(5 \mid 0\).

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Key Concepts

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

Integer Properties
Integers are the set of whole numbers that can be positive, negative, or zero. One of the essential properties of integers is their behavior in division. In our exercise, the expression \( 5 \mid 0 \) asks whether 5 divides 0. This question probes the core property of divisibility. When we say that an integer \( a \) divides another integer \( b \), it means there exists some integer \( k \) such that \( b = a \times k \). For our case, we seek a \( k \) such that \( 0 = 5 \times k \).

This can be true if \( k = 0 \) because multiplying 5 by 0 results in 0. Therefore, 0 can be a product of 5 and another integer, reaffirming that 5 indeed divides 0. Understanding and navigating through such properties is fundamental in solving integer-related problems, providing a clearer understanding of how integers interact with each other via basic operations like division.
Zero in Division
Dealing with zero in division can sometimes be confusing, but it's straightforward once you grasp the underlying logic. Zero, as a number, holds a special place in mathematics. It acts as a neutral element in addition and subtraction, but it also has unique characteristics in multiplication and division. Any number multiplied by zero equals zero, which is essential when considering divisibility.

In our specific question, \( 5 \mid 0 \) translates to determining if 0 can be expressed as \( 5 \times k \) for some integer \( k \). Since this expression holds when \( k = 0 \), it confirms that zero is divisible by any non-zero integer.
  • Zero divided by any non-zero number is always zero: \( 0 \div 5 = 0 \).
  • A number cannot be divided by zero, as this is undefined.
  • Zero being the product of multiplication emphasizes its divisibility nature.
The handling of zero in division thus sheds light on its divisibility rules, which help in solving broader mathematical problems.
Mathematical Reasoning
Mathematical reasoning plays a crucial role in understanding and solving complex problems. It involves the use of logic to evaluate relationships between numbers and determine truths. In this exercise, we used reasoning to check if 5 divides 0. Our reasoning follows logical steps to validate whether there exists an integer \( k \) such that \( 0 = 5 \times k \).

Through logical deduction and application of integer properties, we found that \( k = 0 \) satisfies this equation, confirming 5 divides 0. This process involves understanding definitions, applying mathematical properties, and checking the conditions needed to prove divisibility.
  • Logical deduction helped us find \( k = 0 \) as a potential solution.
  • Applying properties of zero in multiplication and division simplified the reasoning.
  • Validating conditions derived from definitions proved the initial statement, \( 5 \mid 0 \).
Understanding and using sound reasoning enhances analytical skills, making it easier to tackle various mathematical challenges effectively.

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Most popular questions from this chapter

a. Use proof by contradiction to show that for any integer \(n\), it is impossible for \(n\) to equal both \(3 q_{1}+r_{1}\) and \(3 q_{2}+r_{2}\), where \(q_{1}, q_{2}, r_{1}\), and \(r_{2}\), are integers, \(0 \leq r_{1}<\) \(3,0 \leq r_{2}<3\), and \(r_{1} \neq r_{2}\). b. Use proof by contradiction, the quotient-remainder theorem, division into cases, and the result of part (a) to prove that for all integers \(n\), if \(n^{2}\) is divisible by 3 then \(n\) is divisible by 3 . c. Prove that \(\sqrt{3}\) is irrational.

If \(r\) is any rational number and \(s\) is any irrational number, then \(r / s\) is irrational.

If \(c\) is a positive real number and \(x\) is any real number, then \(-c \leq x \leq c\) if, and only if, \(|x| \leq c\). (To prove a statement of the form " \(A\) if, and only if, \(B\)." you must prove "if \(A\) then \(B\) " and "if \(B\) then \(A . "\) )

Fill in the blanks in the following proof by contraposition that for all integers \(n\), if \(5 X n^{2}\) then \(5 X n\). Proof (by contraposition): [The contrapositive is: For all integers \(n\), if \(5 \mid n\) then \(5\left\lfloor n^{2}\right.\).] Suppose \(n\) is any integer such that \(\frac{(\mathrm{a})}{-}\) [We must show that (b) ] By definition of divisibility, \(n=\) (c) for some integer \(k\). By substitution, \(n^{2}=\frac{(\mathrm{d})}{-5\left(5 k^{2}\right), \text { But } 5 k^{2} \text { is an integer because it }}\) is a product of integers. Hence \(n^{2}=5 \cdot\) (an integer), and so (e) \([\) as was to be shown \(]\).

Definition: The least common multiple of two nonzero integers \(a\) and \(b\), denoted \(\operatorname{lcm}(a, b)\), is the positive integer \(c\) such that a. \(a \mid c\) and \(b \mid c\) b. for all integers \(m\), if \(a \mid m\) and \(b \mid m\), then \(c \mid m\). Prove that for all positive integers \(a\) and \(b, \operatorname{gcd}(a, b)=\) \(\operatorname{lcm}(a, b)\) if, and only if \(a=b\).
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