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7 . Fix a positive m > 1 and define complete set of residues mod to be a set of integers S such that every integer is congruent to exactly one integer in the set...

Question

7 . Fix a positive m > 1 and define complete set of residues mod to be a set of integers S such that every integer is congruent to exactly one integer in the set $. For example, by theorem section 4 of Dudley; the set $ {0,1, mn = 1} is a complete set of residues mod m and is usually called the complete set of least residues mod m" Describe a complete set S of residues mod 13 where every I € S satisfies 20 < = < 40. Describe a complete set S of residues mod 13 where every € €

7 . Fix a positive m > 1 and define complete set of residues mod to be a set of integers S such that every integer is congruent to exactly one integer in the set $. For example, by theorem section 4 of Dudley; the set $ {0,1, mn = 1} is a complete set of residues mod m and is usually called the complete set of least residues mod m" Describe a complete set S of residues mod 13 where every I € S satisfies 20 < = < 40. Describe a complete set S of residues mod 13 where every € € S is negative: Describe complete set S of residues mod 13 consisting entirely of odd integers Assume that m > 1 is an odd positive integer: Show m _1 m -3 S = {= m -3 "24} 1-1,0,1, is a complete set of residues mod Explicitly list this complete set of residues for mod 13. Let a > 1, (a,m) = 1 and S is a complete set of residues mod m. Form the new set Sa = {arlz € S}. Prove Sa is complete set of residues mod m. Show by example that the result will fail if (a,m) >1



Answers

Find counterexamples to each of these statements about congruences.
a) If $a c \equiv b c(\bmod m),$ where $a, b, c,$ and $m$ are integers with $m \geq 2,$ then $a \equiv b(\bmod m) .$
b) If $a \equiv b(\bmod m)$ and $c \equiv d(\bmod m),$ where $a, b, c, d,$ and $m$ are integers with $c$ and $d$ positive and $m \geq 2,$ then $a^{c} \equiv b^{d}(\bmod m) .$

It's so in this question, we want to show that a is equivalent to be more em if a more M is going to be so. We first start up with this statement and one produces what is what he's saying. So to use the statement well, first was the first part of it. So if we can, we know that we can write a easy to see m suspect be where? Sea and the Vintages. We also know we can write be as being e m plus by f where e and f r vintages. So by definition, off mods this first month is just beat the second What is this, Steph? So day is going to now If we saw four day we have be easy to a miles by C m. We have f busy with you ever be monster he and so using this equality we now have a month of C M is going to be modest. We'll see now it was attract be across so they must be using. This is a e e plus See America the CM must e m so we can talk it out and we just see my sweet So what we know is that C and E are indigenous. So when we subtract him, this is just another interject here. So this is m tongue. When it does our music to negative a Mars will be so since we have m tired by Inter, what this means is that M is the divisor off a man despite be so now since any devised by a B, by definition, off con fluency eighties equivalents to be more and said this implies this.

Okay, so in question, we want to show that if a is equal to be mobbed em, then a more M is equal to be. Well, if a is equal to be more M, that means that exists at Inter Jacquet such that a is equal to okay and plus might be This is by definition, no more now moving be to one side, equal to minus by K. M. Well, now we haven't been to jail here, So you feel that this being a just by Let's say well and where l is its martyrs k, he's an insurgent. So this is what's again, by definition of modern, we have b is equal to a Madi. Yeah, as required.

So this question. We want to prove that if the greatest common device between end is wetter than one, then the inverse Oh, a mud m doesn't exist. So what we'll do is we'll prove that contradiction and we assume that they exist. This be such that a B, it's equivalent to one mud. And so this statement, he implies two things. Two things. First, what he implies is that a big is equivalent to one just by K Okay, em where Kay is an integer So this is just a simple definition off mother. Any also implies that's a B is what minus one is equivalent to zero mobile. And because this is this is zero implies that M is a divisor off a B minus one. So what do we do with these two things? Well, festival. We know that. Hey, is it divisor all we know? So it will let d to be the greatest common divisor or A and M. We know that day is devices off a and the visor open. So this is just the ignition off his common device. Now, if we look back to this definition, this definition if we move, came to the other side. We have a B minus y k and is equivalent to one. So when we multiply, be to this. We still get Dean would get device off. This which is we should be pretty clear by the A. T. As anything. Day is a divisive home. Kay, as a result is divisor all a b marshall km. And this is one that this is size D is the visor. The problem is the only number that divides one is one, is it? What? But this is a contradiction because we have that d which is equal to the greatest common divisor are A and M must have rather than one. But we have one. This is a contradiction. So then therefore that does not exist this school and hence the investors that exist.

Okay, So for this question, we want to find camera examples to disprove the fact that a is a sea is calm. Going to be C does not imply a fork or to be so to do this, we'll take them as being fine, see as being five a as one and be as to so if if we prove that distinct satisfied so a c is equivalent to one. Or is it 21 times five? So if we take that more M, which is five, he's going to be five. Mud five status zero BC is two times five. You talk about five mull and which is just five. That's 13 10 more. Five said that zero 2000 So therefore decide it's satisfying. But this right side is not satisfied because a easy to work easy. So that is one is not going to be moat flying. So what this question? What will do Upset for Poppy? Sorry. What we'll do is we'll take em as being three a Is it going to be easy to see? Is it into two Empty? Easy to why So the first cases satisfied because she is congruent to more. Three. This one is also satisfied because she is In fact, I'm going to five months. Three just the past treatment. So if we compute this, we have eight. The sea is even too true to the truth. That's four more. Three. So that's one B to the a d. Is it too true to the five which is equal to 32? And if we take that more three. This is minus one or one three or equivalently two. Mom and one is not hungry with Thio. What me Is that all this is implied that


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