A proof of the binomial theorem

Transcription

1 A poof of the binomial theoem If n is a natual numbe, let n! denote the poduct of the numbes,2,3,,n. So! =, 2! = 2 = 2, 3! = 2 3 = 6, 4! = = 24 and so on. We also let 0! =. If n is a non-negative intege and is an intege such that 0 n, then let ( n) = n!!(n )!. In fact ( n ) is the numbe of subsets of size in a set of size n (but we shall not need this fact in the discussion below). Notice that ( ( n 0) = n n) = fo all n 0. Execise. Show that ( n) ( + n ) ( = n+ ) fo all n and all such that n. solution. This is a diect calculation. You can compute as follows: ( ) ( ) n n n! + =!(n )! + n! ( )!(n +)! n! =!(n +)! (n +)+ n!!(n +)! n! = ((n +)+)!(n +)! (n+)! =!(n+ )! ( ) n+ =. Execise 2. Fo all natual numbes n show that n ( ) n () (+x) n = x. solution. We poceed by induction on n. Fo n =, notice that the left hand side of equation () is equal to (+x) = (+x). While, fo n =, the ight hand side of equation () is equal to ( ) ( ) ( ) x = x 0 + x = (+x). 0 =0 So equation () holds fo n =. By induction, suppose equation () holds fo some intege k. Now we calculate =0 (+x) k+ = (+x)(+x) k = (+x) k = k =0 ( k =0 ( k ) x ) x + k =0 ( ) k x + whee the second equality follows fom the induction hypothesis. Isolating out the fist tem fom the fist sum and the final tem fom the second sum, we can ewite the above equation as ( (2) (+x) k+ = + ( ) (( k k k ( ) k )x + )x )+x + k+. = =0

2 Now we shall ewite the second sum in equation (2) by changing the index of summation as follows: ( ) k k x + = ( ) k k x s =0 s= s = ( ) k k x. = Substituting in equation (2), we obtain (+x) k+ = + = + = + ( k ( k ( k = = = ( ) k + = x = ( ) k+ k+ x. =0 ( ) ( k k ( ) k )x + )x = ( (k ) ( ) ) ) k + x +x k+ ( ) k + )x ( k = +x k+ ( ) k+ )x + +x k+ ( ) k+ x k+ k+ The second equality above follows by combining the two sums tem by tem and the thid equality above follows fom execise. Thus we have agued that if equation () holds fo n = k, then equation () also holds fo n = (k +). By induction on n, it follows that equation () holds tue fo all natual numbe n. 2

3 . The eal numbe system Definition.. Let A be a set. A elation < on A is called an ode elation if a < b and b < c implies a < c. fo all a,b A, exactly one of the following is tue: (a < b) o (a = b) o (b < a). Definition.2. Let A be a set and let < be an ode elation on A. We shall wite a b if a < b o a = b. Let E be a subset of A. Say that b A is a uppe bound fo E if x < b fo all x E. Say that E is bounded above if E has an uppe bound. Definition.3. Let A be a set and let < be an ode elation on A. Let E A. We say that b 0 is a least uppe bound o supemum of E if b 0 is an uppe bound of E. if b is any uppe bound of E, then b 0 b. It is easy to see that if b and b ae both least uppe bounds of E, then b = b. If b is the least uppe bound of b, then we wite b = sup(e). Similaly define geatest lowe bound, o infimum, denoted inf(e). Definition.4. A binay opeation m on a set A is a function m : A A A. Often we wite m(a,b) = a b and say call the binay opeation. Fo example: + is a binay opeation on N. is not a binay opeation on N but is a binay opeation on Z. is binay opeation on P(N)..5. Fact: Thee exists a set R with two binay opeations + (plus) and (dot) called addition and multiplication, two special elements called 0(zeo) and (one) and a ode elation < satisfying the following twelve axioms: () (x+y) = (y +x) fo all x,y R. (2) (x+y)+z = x+(y +z) fo all x,y,z R. (3) x+0 = x fo all x R. (4) Fo all x R, thee exists an element, denoted x R such that x+( x) = 0. (5) x.y = y.x fo all x,y R. (6) x.(y.z) = (x.y).z fo all x,y,z R. (7) x. = x fo all x R. (8) Fo all x R {0}, thee exists an element, denoted x R such that x.x =. (9) x.(y +z) = x.y +x.z fo all x,y,z R. (0) Fo all x,y,z R, if x < y, then x+z < y +z. () Fo all x,y R, if x > 0 and y > 0, then x.y > 0. (2) If E is a non-empty subset of R that is bounded above, then E has a least uppe bound. We shall say that a eal numbe x is positive if x > 0 and negative if x < 0. Stating fom (R,+,.,0,,<) satisfying the axioms above, we can stat poving small things like a =.a fo all a R, ( ).( ) =, > 0, If x R {0}, then x 2 > 0, If x,y,z R such that x > 0 and y > z, then xy > xz If E R is bounded below, then it has a geatest lowe bound, i.e. inf(e) exists. and so on. We can identify the set of ational numbes inside R as a subset. Then we can stat developing the basic popeties of eal numbe system and stat poving the theoems of calculus. 3

4 We should emak that the set of ational numbes Q satisfies axioms though but not axiom 2. The failue of axiom 2 is illustated by the following execise: Execise.6. Let S = {x Q: x 2 < 2}. Then show that S is a non-empty subset of Q that is bounded above, yet S does not have a least uppe bound in Q. Execise.7 (See.2.3 in text). Thee exists a positive eal numbe x R such that x 2 = 2. Fo ou pupose the Least uppe bound axiom is the most impotant one. Theoem.8 (Achimedian Popety). Let x,y R with x > 0. Then thee exists n N such that nx > y. Poof. Suppose not. Then S = {nx: n N} is bounded above by y and hence b = sups exists. Then (b x) is not an uppe bound of S, so thee exists n N such that nx > b x, which implies (n+)x > b, which implies b is not an uppe bound of S contadicting b = sups. Theoem.9 (ationals ae dense in eals). If x,y R such that x < y, then thee exists q Q such that x < q < y. Poof. Since (y x) > 0, by Achimedean popety, thee exists n N such that n(y x) >, so ny > nx +. Also by Achimedean popety, thee exists a natual numbe that is geate than nx. So by the Well odeing pinciple of natual numbes, thee exists a smallest natual numbe m such that m > nx. If m > nx +, then that would imply (m ) > nx contadicting the minimality of m, hence m nx+. Now we have nx < m nx+ < ny. So x m n < y. Theoem.0. Let S = {/n: n N}. Then inf(s) = 0. Poof. Note that S is non-empty and S is bounded below by 0. So x = infs exists Since 0 is a lowe bound fo S, we have x 0. If possible suppose x > 0. By achimedean popety thee exists n N such that nx >, so x > /n which contadicts x is a lowe bound fo S. So we must have x 0. It follows that x = 0. When, woking with the supemum of a set we often use the following popety without explicitly mentioning it. The poof is left out as an easy execise. Lemma.. (a) Let E be a nonempty subset of R that is bounded above. Then fo evey ǫ > 0. thee exists an element of E in the inteval (sup(e) ǫ,sup(e)]. (b) Let E be a nonempty subset of R that is bounded below. Then fo evey ǫ > 0. thee exists an element of E in the inteval [inf(e),inf(e)+ǫ). Lemma.2. (a) Let A and B be two subsets of R that ae bounded below. Suppose fo each b B thee exists a A such that a b. Then infa infb. (b) Suppose A is a subset of R that is bounded below. Suppose fo each m N, thee exists a A such that a < /m. Then infa 0. Poof. Let b B. Then thee exists a A such that a b. So infa b. In othe wods infa is a lowe bound fo B. So By definition of infb, we get infa infb. This poves pat (a). Pat (b) follows fom pat (a) with B = {/n: n N} since infb = 0 by the pevious theoem. Execise.3. Let A = { 2 n + n 2 : n N}. Show that infa = 0. sketch of poof. Let m N. Then < 2 4m 2 2m. Note that a = 4m + A and a = 2 4m 2 4m + < 4m 2 2m + 2m < m. No pat (b) of the pevious esult implies infa 0. Also, clealy 0 is a lowe bound of A, so infa 0. Lemma.4. Let E be a nonempty set of eal numbes such that infe = 0. Suppose a b+e fo all e E. Then a b. 4

5 Poof. If possible, suppsose a > b, Then a b > 0. So (a b) cannot be a lowe bound fo E. In othe wods, thee exists e E such that (a b) > e, that is a > b+e which is a contadiction. So we must have a b. Execise.5. Let x,y > 0. If x 2 > y 2, then x > y. Poof. Notice x y because that would imply x 2 = y 2. If possible, suppose x < y. Then x 2 = x.x < x.y < y.y = y 2 which contadicts x 2 > y 2. So we must have x > y. sketch of poof. LetE = {x R: x > 0,x 2 < 2}. ThenE isboundedabove, say by2andnonempty since E. Let b = sup(e). Note that b. If possible, suppose b 2 > 2. Choose n N such that n < 3 (b2 2 ). Then 2 n + 2 n 2 n + n 3 n < b2 2. Reaanging the inequality, we get 2( + n )2 < b 2, so 2 < (b( n (b( n n+ ))2, hence x < b( n n+ ), so b( n n+ n+ ))2. So if x E, then x 2 < ) is an uppe bound fo E which is less than b, contadicting b = supe, so b 2 2. On the othe hand, suppose b 2 < 2. Choose n N such that n < 2 ( b2 2 ). Then 2 n n 2 2 n < b2 2. Reaanging, we get b 2 < 2( n n )2 o that (b( n n ))2 < 2. But this means b < b( n n ) E, which contadicts the assumption that b is an uppe bound fo E. So b 2 2. Next we need to lean basic popeties of woking with supemum and infimum. Fo example sup(x+a) = x+supa fo all x R etc (See pop..2.6,.2.7). Finally we need the basic popeties of the absolute value function (see section.3.,.3.2,.3.3,.3.4), in paticula the tiangle inequality x + y x+y and its vaiant x y + y z x z fo all x,y,z R. To pove the tiangle inequality Note that x x x and y y y and add these togethe 5

6 2. Sequences Definition 2.. A (eal) sequence is a function fom N R. Let a : N R be a sequence. Usually we wite a n = a(n) fo n N and we denote the sequence simply by (a,a 2,,a n, ) o by {a n } n= in shot. Fo example {n} n= denotes the sequence,2,3,, {/n 2 } n= denotes the sequence,/4,/9,/6,, The Fibinacci sequence {f n } is defined inductively by f = f 2 = and f n = f n +f n 2 fo all n > 2. Let {a n } be a sequence. We say that {a n } n= is bounded above if the set {a n: n N} is bounded above et cetea. Definition 2.2. Let {a n } n= be a sequence of eal numbes and a R. Say that lim n a n = a, if given any ǫ R >0, thee exists N N such that a n a < ǫ fo all n > N. We say that a sequence {a n } n= is convegent, if the sequence {a n} n= has a limit. When we say lim n a n = a, we mean that {a n } n= is convegent and a is the limit of this sequence. Lemma 2.3. (a) lim n n = 0. (b) Let c R and define a sequence {a n } n= by a n = c fo all n. (this is called the constant sequence). Then lim n a n = c. Poof. (a). Let ǫ > 0 be a positive eal numbe. By Achimedian popety, thee exists N N such that Nǫ >, so ǫ > /N. Let n N such that n > N. Then n 0 = n < N < ǫ. So by the definition of limit, it follows that { n } n= is convegent and lim n n = 0. (b) Let ǫ > 0 be a positive eal numbe. Let N =. Note that fo all n > N, we have a n c = 0 < ǫ. So by definition of limit, we find that the constant sequence {a n } n= is convegent and lim n a n = c. Lemma 2.4. Let {b n } be a convegent sequence of eal numbes. Let y R. (a) If lim n b n > y, then thee exists N N such that b n > y fo all n > N. (b) If lim n b n < y, then thee exists N N such that b n < y fo all n > N. Poof. Let b = lim n b n. Then b > y. Choose ǫ = (b y)/2. Note that b ǫ = b 2 + y 2 > y since b > y. By definition of limit, thee exists an N N such that b n b < ǫ fo all n > N. So fo all n > N we have b n b > ǫ, so b n > b ǫ > y. This poves pat (a). Pat (b) is poved in simila manne. Execise 2.5. Let {b n } n= be a convegent sequence. Let lim nb n = b. (a) Thee exists N N such that b n < b + fo all n > N. (b) Assume b 0. Then thee exists N N such that b n > b /2 fo all n > N. Poof. (a) Note that b + > b and ( b + ) < b < ( b + ). By lemma (a), we can find N N such that b n > ( b + ) fo all n > N. By lemma (b) we can find N N such that b n < ( b +) fo all n > N 2. Let N = max{n,n 2 }. If n is a natual numbe such that n > N, then ( b + ) < b n < ( b + ), so b n < b +. This poves pat (a). Pat (b) is left as an execise. Lemma 2.6. Let {a n } and {b n } be two convegent sequences. Let a = lima n and b = limb n. Then (a) {a n ±b n } n= is convegent and lim n(a n ±b n ) = a±b. (b) {a n b n } n= is convegent and lim na n b n = ab. (c) If c R, then {cb n } n= is convegent and lim nca n = ca. (d) If b 0, Then {a n /b n } n= is convegent and lima n/b n = a/b. 6

7 ǫ Poof. (b) Choose ǫ > 0. Let δ = a + b +. Note that δ > 0. By the execise above, thee exists N N such that a n ( a +) fo all n N. By definition of limit, thee exists N 2 N such that a n a δ fo all n N 2. By definition of limit, thee exists N 3 N such that b n b δ fo all n N 3. Choose a natual numbe N such that N max{n,n 2,N 3 }. Let n N. Then a n b n ab a n b n b + b a n a ( a +)δ + b δ = ǫ (c) Let {b n } be the constant sequence: b n = c fo all n. Then limb n = c. Now pat (c) follows fom pat (b). Theoem 2.7. Show that thee exists x R such that x 2 = 2. Poof. Let S = {x R: x 2 < 2,x > 0}. Note that S, so S. Also if x S and x > 2, then that would imply x 2 > 4 > 2 which is impossible. So x S implies x 2, that is S is bounded above. So by completeness axiom sups exists. Let b = sups. Since x S, we find that b. Suppose b 2 < 2. Conside the sequence b n = (b+ n )2. Note that lim n b n = b 2 < 2. So by lemma 2.4, thee exists n N such that b n < 2 fo some n. In othe wods (b+ n )2 < 2 fo some n which means (b+ n ) S which contadicts the fact that b is an uppe bound fo S. It follows that b2 2. Now suppose b 2 > 2. Then conside the sequence c n = (b n )2. Note that lim)nc n = b 2 > 2. So by lemma 2.4, thee exists n NN such that c n > 2. So if x S, then x 2 < 2, so x 2 < (b n )2. Since b >, we know (b n ) > 0. Thus it follows that x < (b /n). This means (b n ) is an uppe bound of S. But this contadicts the fact that b is the least uppe bound of S. So b

8 3. A few execises Execise 3.. Let B A R with A is bounded above and x = supa. Suppose fo all a A, thee exists b B such that b a. Then show that supb exists and supb = x. solution. If b B, then b A (since B A), so b supa = x. So x is an uppe bound fo B. So by completeness axiom supb exists. Since x is an uppe bound fo B, we have supb x. Now let y be any eal numbe such that y < x. Then y is not an uppe bound fo A, so thee exists a A such that y < a. By given condition, thee exists b B such that b a. So y < b. Thus we find that if y < x, then y is not an uppe bound fo B. In othe wods, any uppe bound fo B is geate than o equal to x. Since we aleady know that x is an uppe bound fo B, it follows that x is the least uppe bound fo B, that is, supb = x. Execise 3.2. Let a,b be two eal numbes. Suppose that a ǫ < b fo all eal numbes ǫ > 0. Then show that a b. Poof. If possible, suppose a > b. Choose ǫ = (a b)/2. Since a > b, we have ǫ > 0, which implies a b = 2ǫ > ǫ. So a ǫ > b, and this contadictions ou assumption. Thus we must have a b. Execise 3.3. Let A,B be two bounded subsets of R. Define C = {a+b: a A,b B}. Show that sup(c) exists and sup(c) = sup(a) + sup(b). Poof. Since A and B ae nonempty subsets of R and bounded above, By completeness axiom sup(a) and sup(b) exists. Let c C. Then thee exists a A and b B such that c = a+b. Now a sup(a) and b sup(b). So c sup(a)+sup(b). So sup(a)+sup(b) is an uppe bound fo the set C. By completeness axiom, sup(c) exists and we also get sup(c) sup(a)+sup(b). Now let ǫ > 0 be any positive eal numbe. Then sup(a) ǫ 2 is not an uppe bound of A since sup(a) ǫ 2 < sup(a). So thee exists a A such that a > sup(a) ǫ 2. By simila easoning, thee exists b B such that b > sup(b) ǫ 2. Since a + b C, we have a + b sup(c). So So sup(a)+sup(b) ǫ < a+b sup(c). Thus sup(a)+sup(b) ǫ < sup(c) fo any ǫ > 0. By the pevious execise, it follows that sup(a) + sup(b) sup(c). Combining the esults of the pevious two paagaphs, we find that sup(a)+sup(b)+sup(c). Execise 3.4. Let p be a pime numbe and let a,b Z such that p divides ab. Then show that p divides a o p divides b. Poof. Suppose p does not divide a. Since p is a pime numbe, the only factos of p ae and p. So it follows that gcd(a,p) =. Then we know that thee exists,s Z such that a + ps =. Now notice that b = ab +bps. Note that p divides ab (since p divides ab) and p divides bps. It follows that p divides b. Thus we have agued that if p does not divide a then p divides b. So p divides a o b. Execise 3.5. Let f : X Y be a function. Let A,A 2 X. Show that f(a A 2 ) f(a ) f(a 2 ). Show that equality need not hold. Poof. Let b f(a A 2 ). Then we can wite b = f(a) fo some a A A 2. Since a A, we have b = f(a) f(a ). Since a A 2, we have b = f(a) f(a 2 ). Thus b f(a ) f(a 2 ). Thus we have agued that f(a A 2 ) f(a ) f(a 2 ). To see that we need not have f(a A 2 ) = f(a ) f(a 2 ) conside the following example: Let X = Y = {,2} and f : X Y be the function defined by f() = f(2) =. Let A = {}, A 2 = {2}. Then we note that A A 2 = and hence f(a A 2 ) =. On the othe hand f(a ) = {} = f(a 2 ) and hence f(a ) f(a 2 ) = {}. So in this example f(a A 2 ) is a pope subset of f(a ) f(a 2 ). 8

9 Execise 3.6. Show that (3) fo all n N. 2 n j= j + n 2. solution. Fo n =, we have 2 j= j = + 2. So the inequality (3) holds fo n =. By induction, suppose the inequality holds fo some m N. By induction hypothesis we have 2 m j= j + m 2. Now conside the sum 2 m+ j=2 m + j. Note that this sum has 2m tems and each summand is at least /2m+. Thus we obtain 2 m+ j=2 m + j 2m. = 2 m+ 2. By adding the above inequalities, it follows that 2 m+ j= j = 2 m j= j + 2 m+ j=2 m + j (+ m 2 )+ 2 = + m+ 2. Thus we have agued that if the inequality (3) holds tue fo some m N, the it also holds tue fo m+. By induction on n, it follows that (3) holds tue fo all n N. Execise 3.7. Let a n = 2 n j= j fo n N. Show that lim na n does not exist. Poof. If possible, supposelim n a n = a. Then theeexists N N such that a n < a+ fo all n > N. By achimedean popety, we can find a natual numbe m such that m > 2a, so + m 2 > a+, and so a m > a+. But since a m a m+ a m+2, it follows that a n > a+ fo all n > m, which is a contadiction. This poves lim n a n does not exist. Execise 3.8. Let x < y be two eal numbes. Show that thee is an iational numbe a such that x < a < y. You may use the fact that thee is a eal numbe 2 > 0 such that ( 2) 2 = 2. Poof. Fist we shall pove that 2 / Q. We shall ague by contadiction. So suppose 2 Q. Then we can wite 2 = m /n whee m,n Z and n 0. Let d = gcd(m,n ). Wite m = m /d and n = n /d, so that gcd(m,n) =. Note that 2 = m /n = m/n. Squaing both sides, we get 20 = m 2 /n 2, so n 2 = 2m 2. It follows that 2 n 2. But since 2 is a pime, this implies 2 n. So n = 2k fo some k Z. So 2m 2 = n 2 = (2k) 2 = 4k 2 and hence m 2 = 2k 2. But this implies 2 m 2 and hence 2 m. Thus we find 2 divides both m and n which contadicts gcd(m,n) =. This contadiction poves that b / Q. Now conside the inteval (x 2,y 2). We know thee exists a ational numbe q such that x 2 < q < y 2. Then x < q 2 < y. Let a = q/ 2. If a Q, then we would have 2 = q/a Q as well which would be a contadiction. So a / Q and x < a < y. Execise 3.9. Let S be an infinite subset of (0,) such that fo each n N, thee exists only finitely many elements of S that ae geate than /n. Show that inf(s) = 0. Poof. We ae told that S is inifinite; in paticula, S is nonempty. Since S (0,), we have x > 0 fo all x S. Since S is nonempty and bounded below, inf(s) exists by the completeness axiom. Now suppose b > 0 be any positive eal numbe. By achimedean popety, we can find a natual numbe n such that nb >, so /n < b. By ou given assumption thee ae only finitely many 9

10 elements of S that ae geate than /n. Since S is infinite, it follows that thee exists x S such that x < /n. Then x < b. Thus we have agued that if b > 0, then b cannot be a lowe bound fo S. In othe wods, if b is a lowe bound fo S, then b 0. Note that 0 is a lowe bound fo S since S (0,). It follows that 0 must be the geatest lowe bound fo S. 0