Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Exercise 5.2.4. Fill in the blanks in the following proof that the remainder is always unique. We'll give a proof by contradiction. Suppose that a has two different re- mainders when divided by m. Let's call these two different remainders r and s, where 0 < r, s<_<1> and r + 8. It follows that a = q • m+r and a = p·m+ < 2> _, where q and p are <3>. Setting these two expressions equal and rearranging enables us to obtain an expression for r – s, namely: r - s = (<4> )•m. Thus r- s is an integer multiple of <5> . On the other hand, we know that r 2 0 and s < _<6>, so by arithmetic we obtain r - s 2 <7>. Furthermore, r< <8>_ and s2 <9>, so r - s<< 10 > . Combining these two results, we find that r - s is an integer between< 1l> and <12 >. Now, the only integer multiple of m between < 13 > and <14 > is < 15 >. It follows that r – s = < 16 > , or r = <17> . But this contradicts our supposition that _< 18 > _. So our supposition cannot be true: and a cannot have < 19 >. Thus the remainder when a is divided by m is unique, and the proof is complete.

Exercise 5.2.4. Fill in the blanks in the following proof that the remainder is always unique. We'll give a proof by contradiction. Suppose that a has two different re- mainders when divided by m. Let's call these two different remainders r and s, where 0 < r, s<_<1> and r + 8. It follows that a = q • m+r and a = p·m+ < 2> _, where q and p are <3>. Setting these two expressions equal and rearranging enables us to obtain an expression for r – s, namely: r - s = (<4> )•m. Thus r- s is an integer multiple of <5> . On the other hand, we know that r 2 0 and s < _<6>, so by arithmetic we obtain r - s 2 <7>. Furthermore, r< <8>_ and s2 <9>, so r - s<< 10 > . Combining these two results, we find that r - s is an integer between< 1l> and <12 >. Now, the only integer multiple of m between < 13 > and <14 > is < 15 >. It follows that r – s = < 16 > , or r = <17> . But this contradicts our supposition that _< 18 > _. So our supposition cannot be true: and a cannot have < 19 >. Thus the remainder when a is divided by m is unique, and the proof is complete.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question

Please do exercise 5.2.4

Proposition 5.2.3.(The division algorithm) Given any integer a and any positive integer m, then there exists a unique number r between 0 and m - 1 such that a = q - m +r for some integer q. In this expression, q is called the quotient, and r is called the remainder. PROOF. It turns out that proving this "simple" fact is not so simple! Al- though this fact has been used for millennia (it's sometimes called Eu- clidean division, because Euclid used it ca. 300 B.C.), the first rigorous proof was found relatively recently. There are actually two things to prove: first, that the remainder r exists, and second, that it's unique. We're going to punt on the 'existence' part: you can find the proof in a book on num- ber theory.? The 'unique' part is proved by the following fill-in-the-blanks exercise: Exercise 5.2.4. Fill in the blanks in the following proof that the remainder is always unique. We'll give a proof by contradiction. Suppose that a has two different re- mainders when divided by m. Let's call these two different remainders r and s, where 0 < r, s< <1>_ and r + s. It follows that a = q • m+r and a = p · m + <2> , where q and p are <3>. Setting these two expressions equal and rearranging enables us to obtain an expression for r – s, namely: r – s = <4>_) · m. Thus r- s is an integer multiple of < 5 > . On the other hand, we know that r > 0 and s < <6>, so by arithmetic we obtain r – s > <7> . Furthermore, r < <8> _ and s > <9>, so r - s< < 10 > Combining these two results, we find that r - s is an integer between < 11 > _ and <12 > . Now, the only integer multiple of m between < 13 > and <14 > is < 15 >. It follows that r – s = < 16 > _, or r = < 17 > _. But this contradicts our supposition that < 18 >. So our supposition cannot be true: and a cannot have < 19 >. Thus the remainder when a is divided by m is unique, and the proof is complete.
Transcribed Image Text:Proposition 5.2.3.(The division algorithm) Given any integer a and any positive integer m, then there exists a unique number r between 0 and m - 1 such that a = q - m +r for some integer q. In this expression, q is called the quotient, and r is called the remainder. PROOF. It turns out that proving this "simple" fact is not so simple! Al- though this fact has been used for millennia (it's sometimes called Eu- clidean division, because Euclid used it ca. 300 B.C.), the first rigorous proof was found relatively recently. There are actually two things to prove: first, that the remainder r exists, and second, that it's unique. We're going to punt on the 'existence' part: you can find the proof in a book on num- ber theory.? The 'unique' part is proved by the following fill-in-the-blanks exercise: Exercise 5.2.4. Fill in the blanks in the following proof that the remainder is always unique. We'll give a proof by contradiction. Suppose that a has two different re- mainders when divided by m. Let's call these two different remainders r and s, where 0 < r, s< <1>_ and r + s. It follows that a = q • m+r and a = p · m + <2> , where q and p are <3>. Setting these two expressions equal and rearranging enables us to obtain an expression for r – s, namely: r – s = <4>_) · m. Thus r- s is an integer multiple of < 5 > . On the other hand, we know that r > 0 and s < <6>, so by arithmetic we obtain r – s > <7> . Furthermore, r < <8> _ and s > <9>, so r - s< < 10 > Combining these two results, we find that r - s is an integer between < 11 > _ and <12 > . Now, the only integer multiple of m between < 13 > and <14 > is < 15 >. It follows that r – s = < 16 > _, or r = < 17 > _. But this contradicts our supposition that < 18 >. So our supposition cannot be true: and a cannot have < 19 >. Thus the remainder when a is divided by m is unique, and the proof is complete.
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS