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Suppose a # b (mod m). Let r be the remainder of a when divided by <1> , and let s be the remainder of < 2 > _when divided by <3 > . Since the remainders are unequal, it follows that one must be bigger than the other: let us choose a to be the number with the larger remainder, so that r > <4> . By the definition of remainder, we may write a = p · m+ < 5> , and we may also write b = q. < 6> + <7> . Then by basic algebra, a– b = (p – q) ·<8 > + (r- <9> ). We want to show that r - s is the remainder of a – b when divided by m. To do this, we need to show that r - s is between 0 and < 10 > . Since r > s it follows that r – s > <11> . Furthermore, Since r 0, it follows that r-s< _ < 12 > . So we have shown that r - s is between < 13 > and<14> that r- s is the remainder of a – b when divided by m. However, r – s> 0, which means that a – b is not divisible by_ < 16 >. This is exactly what we needed to prove, so the proof is complete. , so by Proposition <15 > _ it follows

Suppose a # b (mod m). Let r be the remainder of a when divided by <1> , and let s be the remainder of < 2 > _when divided by <3 > . Since the remainders are unequal, it follows that one must be bigger than the other: let us choose a to be the number with the larger remainder, so that r > <4> . By the definition of remainder, we may write a = p · m+ < 5> , and we may also write b = q. < 6> + <7> . Then by basic algebra, a– b = (p – q) ·<8 > + (r- <9> ). We want to show that r - s is the remainder of a – b when divided by m. To do this, we need to show that r - s is between 0 and < 10 > . Since r > s it follows that r – s > <11> . Furthermore, Since r 0, it follows that r-s< _ < 12 > . So we have shown that r - s is between < 13 > and<14> that r- s is the remainder of a – b when divided by m. However, r – s> 0, which means that a – b is not divisible by_ < 16 >. This is exactly what we needed to prove, so the proof is complete. , so by Proposition <15 > _ it follows

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
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Please do Exercise 5.2.9

Suppose a # b (mod m). Let r be the remainder of a when divided by <1> , and let s be the remainder of < 2 > _when divided by <3 > . Since the remainders are unequal, it follows that one must be bigger than the other: let us choose a to be the number with the larger remainder, so that r > <4> . By the definition of remainder, we may write a = p · m+ < 5> , and we may also write b = q. < 6> + <7> . Then by basic algebra, a– b = (p – q) ·<8 > + (r- <9> ). We want to show that r - s is the remainder of a – b when divided by m. To do this, we need to show that r - s is between 0 and < 10 > . Since r > s it follows that r – s > <11> . Furthermore, Since r <m and s > 0, it follows that r-s< _ < 12 > . So we have shown that r - s is between < 13 > and<14> that r- s is the remainder of a – b when divided by m. However, r – s> 0, which means that a – b is not divisible by_ < 16 >. This is exactly what we needed to prove, so the proof is complete. , so by Proposition <15 > _ it follows
Transcribed Image Text:Suppose a # b (mod m). Let r be the remainder of a when divided by <1> , and let s be the remainder of < 2 > _when divided by <3 > . Since the remainders are unequal, it follows that one must be bigger than the other: let us choose a to be the number with the larger remainder, so that r > <4> . By the definition of remainder, we may write a = p · m+ < 5> , and we may also write b = q. < 6> + <7> . Then by basic algebra, a– b = (p – q) ·<8 > + (r- <9> ). We want to show that r - s is the remainder of a – b when divided by m. To do this, we need to show that r - s is between 0 and < 10 > . Since r > s it follows that r – s > <11> . Furthermore, Since r <m and s > 0, it follows that r-s< _ < 12 > . So we have shown that r - s is between < 13 > and<14> that r- s is the remainder of a – b when divided by m. However, r – s> 0, which means that a – b is not divisible by_ < 16 >. This is exactly what we needed to prove, so the proof is complete. , so by Proposition <15 > _ it follows
PROOF. Given that a =r (mod m), by the definition of modular equiva- lence it follows that a and r have the same remainder mod m. But since 0<r<m – 1, the remainder of r is r itself. It follows that the remainder of a is also r: so r = mod(a, m). There is an alternative (and very useful) way to determine modular equivalence. Suppose that a = b (mod m), so that a and b have the same remainder when divided by m. Let's call this remainder r. Then we can write a = p- m | r and b = q - m | r for some integers p, q It follows from basic algebra that a – p- m = b-q · m. We then proceed step-by-step using basic algebra as follows: e a - p. m = b– q · m + a- b= p.m - q. m а -b3 (р-4) - т. + a - b is divisible by m. In summary, we have shown that If a = b (mod m) then a – b is divisible by m. which we can also write as a =b (mod m) = a - b is divisible by m. It turns out that the converse statement is also true. The converse state- ment is: If a – b is divisible by m then a = b (mod m). One way to prove this is to prove the contrapositive, which is logically equiv- alent. In this case, the contrapositive statement is, "If a # b (mod m), then a - b is not divisible by m"). Exercise 5.2.9. Finish the proof of the contrapositive by filling in the blanks:
Transcribed Image Text:PROOF. Given that a =r (mod m), by the definition of modular equiva- lence it follows that a and r have the same remainder mod m. But since 0<r<m – 1, the remainder of r is r itself. It follows that the remainder of a is also r: so r = mod(a, m). There is an alternative (and very useful) way to determine modular equivalence. Suppose that a = b (mod m), so that a and b have the same remainder when divided by m. Let's call this remainder r. Then we can write a = p- m | r and b = q - m | r for some integers p, q It follows from basic algebra that a – p- m = b-q · m. We then proceed step-by-step using basic algebra as follows: e a - p. m = b– q · m + a- b= p.m - q. m а -b3 (р-4) - т. + a - b is divisible by m. In summary, we have shown that If a = b (mod m) then a – b is divisible by m. which we can also write as a =b (mod m) = a - b is divisible by m. It turns out that the converse statement is also true. The converse state- ment is: If a – b is divisible by m then a = b (mod m). One way to prove this is to prove the contrapositive, which is logically equiv- alent. In this case, the contrapositive statement is, "If a # b (mod m), then a - b is not divisible by m"). Exercise 5.2.9. Finish the proof of the contrapositive by filling in the blanks:
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