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 well-defined, 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 well-defined 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 well-defined, 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 top-left-to-bottom-right 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 non-zero 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 non-zero 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 well-defined. 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 Well-Ordering 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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