Math 4310 Solutions to homework 1 Due 9/1/16


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1 Mth 4310 Solutions to homework 1 Due 9/1/16 1. Use the Eucliden lgorithm to find the following gretest common divisors. () gcd(252, 180) = 36 (b) gcd(513, 187) = 1 (c) gcd(7684, 4148) = = = = = = = = = = = = = = = = = = Find the multiplictive inverses of the given elements, or explin why there is not one. () [38] 1 = [59] in Z 83 First we do the Eucliden lgorithm: Substituting bckwrds, Hence [38] 1 = [ 24] = [59] in Z = = = = = 7 (38 7 5) 2 = = ( ) =
2 Mth 4310 (Fll 2016) Solution 1 2 (b) [351] hs no inverse in Z 6669 becuse [351] [19] = [0]. (c) [91] 1 = [451] in Z 2565 First we do the Eucliden lgorithm: Substituting bckwrds, 2565 = = = = = = 6 (17 6 2) 1 = = ( ) = = 91 3 ( ) 16 = Hence [91] 1 = [451] in Z Prove the following fcts out divisors. In the following,, b, c N. () If b, then b c. Since b, there is nturl number k such tht bk =. Multipliction is welldefined, so (bk)c = c. Multipliction is lso ssocitive, so b(kc) = c. Finlly, since N is closed under multipliction, kc N, nd by definition, b c. (b) If b nd c b, then c. Since c b, there is nturl number k 1 such tht Since b, there is nturl number k 2 such tht ck 1 = b. (1) bk 2 =. (2) We cn multiply both sides of (1) by k 2 to get (ck 1 )k 2 = bk 2. Multipliction is ssocitive, so c(k 1 k 2 ) = bk 2. Equlity is trnsitive, so using (2), we hve c(k 1 k 2 ) =. Finlly, N is closed under multipliction, so k 1 k 2 N, nd by definition c. (c) If c nd c b, then c (m + nb) for ny integers m, n. Since c nd c b, there re nturl numbers k 1 nd k 2 such tht ck 1 =, ck 2 = b. Multipliction is welldefined nd ssocitive, so c(k 1 m) = m, c(k 2 n) = bn. Multipliction is lso commuttive, so c(k 1 m) = m, c(k 2 n) = nb.
3 Mth 4310 (Fll 2016) Solution 1 3 Addition is welldefined, so Distributivity gives c(k 1 m) + c(k 2 n) = m + nb. c(k 1 m + k 2 n) = m + nb. Finlly N is closed under multipliction nd ddition, so k 1 m + k 2 n N, nd by definition c (m + nb). 4. Fill in the ddition nd multipliction tles for F 5. + mod mod () How cn we conclude from the ddition nd multipliction tles tht the opertions re commuttive? The tles re symmetric cross the toplefttobottomright digonl, so the opertions re commuttive. (b) How cn we conclude tht 0 is n dditive identity? The first row of the ddition tle shows tht 0 + = for ll, while the first column of the sme tle shows tht + 0 = for ll. This suffices to show tht 0 is n dditive identity. (c) How cn we conclude tht every nonzero element hs multiplictive inverse? Excluding the 0 row, ech row in the multipliction tle hs 1, showing tht for every, there is b such tht = 1. Excluding the 0 column, ech column in the multipliction tle hs 1, showing tht for every, there is b such tht b = 1. Hence every nonzero element hs left nd right multiplictive inverse. (d) Working with elements in F 5, find the solution to the liner eqution 3x + 2 = 4. Adding 3 to both sides, we get 3x = 2. Multiplying both sides by 2, we get x = Tking for grnted tht R is field, we cn verify the field xioms for C. Recll tht for two complex numbers z = + bi nd w = c + di, with, b, c, d R, we hve z + w = ( + c) + (b + d)i nd z w = (c bd) + (d + bc)i. () Prove tht + is ssocitive. (Sorry, it s messy! You hve to do this kind of thing once. Also, ssocitivity of is worse.)
4 Mth 4310 (Fll 2016) Solution 1 4 Our three complex numbers will be z k = k + b k i for k = 1, 2, 3. Using the definition of ddition, (z 1 + z 2 ) + z 3 = (( 1 + b 1 i) + ( 2 + b 2 i)) + ( 3 + b 3 i) = (( ) + (b 1 + b 2 )i) + ( 3 + b 3 i) = (( ) + 3 ) + ((b 1 + b 2 ) + b 3 )i Since ddition for R is ssocitive, the lst line becomes (( ) + 3 ) + ((b 1 + b 2 ) + b 3 )i = ( 1 + ( )) + (b 1 + (b 2 + b 3 ))i. Use the definition of ddition gin but now in the opposite direction: s desired. (z 1 + z 2 ) + z 3 = ( 1 + ( )) + (b 1 + (b 2 + b 3 ))i = ( 1 + b 1 i) + (( ) + (b 2 + b 3 )i) = ( 1 + b 1 i) + (( 2 + b 2 i) + ( 3 + b 3 i)) = z 1 + (z 2 + z 3 ), (b) Suppose z = + bi stisfies 0 nd b 0. Wht is the multiplictive inverse of z? Let y = b i. Becuse nd b re both nonzero, their squres re both 2 +b 2 2 +b 2 positive, so 2 + b 2 > 0. Thus y is welldefined. Now we check the products: ( ) zy = ( + bi) 2 + b 2 b 2 + b 2 i = 2 + b 2 b 2 + b 2 i 2 + b 2 i2. Becuse i 2 = 1 nd multipliction is commuttive in R, this becomes zy = 2 + b b 2 = This shows tht y is right inverse of z. We check the other: ( ) yz = 2 + b 2 b 2 + b 2 i ( + bi) = 2 + b b 2 i b 2 + b 2 i 2 + b b 2 i2. Becuse i 2 = 1 nd multipliction is commuttive in R, this becomes yz = 2 + b b 2 i 2 + b 2 = 2 + b b 2 = b 2 = 1. This shows tht y is left inverse of z. Thus y is the multiplictive inverse of z.
5 Mth 4310 (Fll 2016) Solution 1 5 (c) In prt (b), if, b Q, is z 1 Q[i]? Becuse Q is closed under multipliction nd ddition, if, b re in Q, then so is 2 +b 2. As rgued before, 2 +b 2 0, nd Q is field, so ( 2 +b 2 ) 1 b Q. Then, Q, 2 +b 2 2 +b 2 so z 1 = 2 +b 2 + b 2 +b 2 i is in Q[i]. Extended Glossry. Plese give definition of prime number. Then give n exmple of prime number, n exmple of number tht is not prime number (don t forget to explin why!), nd stte nd prove theorem out prime numbers. In studying multipliction in the nturl numbers, we quickly run into the ide of expressing numbers s products of smller numbers. Since 1 is the multiplictive identity, it is most interesting when the components re not 1. For instnce 12 = 4 3, where both 4 nd 3 re smller thn 6 nd neither is 1. Eventully we cnnot brek numbers down further, giving us our miniml units of multipliction, which we cll prime numbers. Definition 1. A prime number is nturl number greter thn 1 with no positive divisors except 1 nd itself. Exmple 2. 3 is prime number. Any divisors of 3 must be smller thn 3, which mens the only possible nontrivil (i.e. neither 1 nor 3) divisor is 2. However 2 does not divide 3, so 3 is prime. Exmple is not prime becuse 12 is divisible by 2. A nturl question is, how mny prime numbers re there? We will show tht there re infinitely mny of them, but before we cn do tht, we need the following lemm: Lemm 4. Every nturl number greter thn 1 hs prime divisor. Proof. Let S be the set of nturl numbers greter thn 1 which hve no prime divisors, nd suppose S is nonempty. Then by the WellOrdering Principle, S hs lest element l. Since l hs no prime divisors, in prticulr l is not prime number. We defined S such tht l > 1, so l hs divisor which is neither 1 nor itself, which we will cll d. We hve tht d divides l but 1 < d < l, so d must hve prime divisor p, since it is not in S. However if p divides d nd d divides l, then p divides l, which contrdicts our construction of l. Hence our initil ssumption must be incorrect, nd S is empty. Therefore every nturl number greter thn 1 hs prime divisor. Theorem 5. There re infinitely mny prime numbers in N. Proof. Suppose there re only finitely mny prime numbers in N, nd they re 1, 2,..., n. Let b = 1 2 n + 1. Since N is closed under ddition nd multipliction, b is in N. By lemm 4, every nturl number greter thn 1 hs prime divisor, so b hs prime divisor, which must be i for some i. Since i divides 1 2 n nd i divides 1 2 n + 1, i must divide 1. However 1 hs no prime divisors, so our initil ssumption ws wrong, nd there re infinitely mny primes in N.
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