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Chapter 21: Problem 4

Prove the following, where \(\mathrm{k}, \mathrm{m}, \mathrm{n}, \mathrm{q}\), and \(\mathrm{r}\) designate integers. In Theorem 3 , suppose \(m=n q_{1}+r_{1}=n q_{2}+r_{2}\) where \(0 \leq r_{1}, r_{2}

Short Answer

Expert verified
The remainders are equal; thus, \( r_1 = r_2 \).

Step by step solution

01

Understand the Problem Statement

We are given that two expressions represent the same integer \( m \) in terms of integer divisions and remainders: \( m = nq_1 + r_1 \) and \( m = nq_2 + r_2 \), where both remainders satisfy \( 0 \leq r_1, r_2 < n \). We need to show that the difference \( r_1 - r_2 = 0 \).
02

Establish the Equation from the Given Expressions

Since both expressions are equal to \( m \), set them equal to each other: \( nq_1 + r_1 = nq_2 + r_2 \).
03

Rearrange to Isolate Terms Involving Remainders

From the equation \( nq_1 + r_1 = nq_2 + r_2 \), subtract \( nq_2 \) and \( r_2 \) from both sides to obtain: \( r_1 - r_2 = n(q_2 - q_1) \).
04

Analyze the Range of Remainders

Remember that by definition, both \( r_1 \) and \( r_2 \) are between 0 and \( n-1 \), inclusive. Thus, \( 0 \leq r_1, r_2 < n \).
05

Conclude the Proof by Considering Implications of Remainder Differences

Since \( r_1 - r_2 = n(q_2 - q_1) \) and \( r_1 - r_2 \) must also fall between \( -(n-1) \) and \( n-1 \) (as the difference between two integers each less than \( n \)), the only possible integer multiple of \( n \) within this range is 0. Therefore, \( r_1 - r_2 = 0 \).
06

Verify Conclusion with the Remainder Condition

Since the range condition \( -(n-1) \leq 0 \leq n-1 \) holds, and \( (r_1 - r_2 = 0) \) implies that the remainders must be equal, the proof is complete.

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Key Concepts

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

Integer Division
Integer division is a fundamental concept in mathematics where a division of integers results in a quotient and a remainder. When one integer divides another, the division can be expressed as:
  • dividend = divisor 脳 quotient + remainder
For example, if you divide 17 by 5, the result is 3 with a remainder of 2, because 5 goes into 17 three times, with 2 left over.

This can be expressed as:
  • 17 = 5 脳 3 + 2
In the exercise, this principle is reflected in expressions such as \( m = nq_1 + r_1 \). Here, \( m \) is the integer being divided by \( n \), \( q_1 \) is the quotient, and \( r_1 \) is the remainder.
Remainders
Remainders are what is left after division when the dividend isn't evenly divisible by the divisor.
In mathematical terms, if \( m = nq + r \) with \( 0 \leq r < n \), \( r \) is the remainder that ensures any multiple of \( n \) plus \( r \) equals the original number, \( m \).

In the proof exercise, understanding remainders is crucial.
  • The condition \( 0 \leq r_1, r_2 < n \) assures us that both \( r_1 \) and \( r_2 \) fit within a specific range.
  • This range limits our possible non-zero values when comparing the difference \( r_1 - r_2 \).
The concept of remainders ensures that \( r_1 \) and \( r_2 \) are always non-negative and less than \( n \), contributing to the proof that their difference must be zero if \( r_1 - r_2 = n(q_2 - q_1) \).
Mathematical Proof
A mathematical proof is a logical argument that verifies the truth of a mathematical statement. In this exercise, we use a specific type of proof by equating two similar expressions and then isolating variables to show equality.

Steps in the proof:
  • Equalize the two expressions representing integer division with remainders: \( nq_1 + r_1 = nq_2 + r_2 \).
  • Rearrange the equation to focus on the remainder terms: \( r_1 - r_2 = n(q_2 - q_1) \).
  • Since the difference \( r_1 - r_2 \) must be within \( -(n-1) \) and \( n-1 \), it ensures that the only possible integer multiple of \( n \) that fits is zero.
These steps form a concise proof showing that \( r_1 = r_2 \), solely by using logical deduction and manipulation of expressions.
Theoretical Mathematics
Theoretical mathematics involves abstract concepts and in-depth analysis of mathematical truths without necessarily focusing on practical applications.

In this exercise, theoretical math is employed to prove a conjectural statement about integer division:
  • The proof revolves around abstract mathematical logic rather than specific numerical examples.
  • It reflects broader mathematical principles, illustrating how certain properties hold universally.
By dealing with integers, divisors, and remainders, the proof exemplifies the beauty of theoretical mathematics鈥攊t shows how mathematical ideas can confirm truths across many cases, providing a structured way to validate concepts.

In conclusion, the proof is not just about computation but understanding the inherent properties and relationships in mathematics.

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Most popular questions from this chapter

In any ordered integral domain, define \(|a|\) by $$ |a|=\left\\{\begin{array}{rll} a & \text { if } & a \geq 0 \\ -a & \text { if } & a<0 \end{array}\right. $$ Using this definition, prove the following: $$ |a|-|b| \leq|a-b| $$

Let \(A\) be an ordered integral domain. Prove the following, for all \(a, b\), and \(c\) in \(A\). $$ a^{2}+b^{2}>a b, \text { if } a \neq b $$

In any ordered integral domain, define \(|a|\) by $$ |a|=\left\\{\begin{array}{rll} a & \text { if } & a \geq 0 \\ -a & \text { if } & a<0 \end{array}\right. $$ Using this definition, prove the following: $$ a \leq|a| $$

In any ordered integral domain, define \(|a|\) by $$ |a|=\left\\{\begin{array}{rll} a & \text { if } & a \geq 0 \\ -a & \text { if } & a<0 \end{array}\right. $$ Using this definition, prove the following: $$ \text { If } b>0,|a| \leq b \text { iff }-b \leq a \leq b $$

The purpose of this exercise is to give rigorous proofs (using induction) of the basic identities involved in the use of exponents or multiples. If \(A\) is a ring and \(a \in A\), we define \(\mathrm{n} \cdot a\) (where \(\mathrm{n}\) is any positive integer) by the pair of conditions: (i) \(1 \cdot a=a, \quad\) and (ii) \((\mathrm{n}+1) \cdot a=\mathrm{n} \cdot a+a\) Use mathematical induction (with the above definition) to prove that the following are true for all positive integers \(\mathrm{n}\) and all elements \(a, b \in A\) : $$ \mathrm{m} \cdot(\mathrm{n} \cdot a)=(\mathrm{mn}) \cdot a $$
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