Question

Prove the Binomial Theorem, that is Exercises 173 (vi) x+y y for all n e N C) Recall that for all 0rS L is divisible by 8 when n is an odd natural number vii))Show that 2 (vin) Prove Leibniz's Theorem for repeated differentiation of a product: If ande are functions of x, then prove that d (uv) d + +Mat0 for all n e N, where u, and d'a d/v and dy da respectively denote (You will need to use but proving this identity is true should not pose much difficulty.) Proνebn Σ-Σ) (ix) is an integer for all 0 srsn Show that (x) Let X be a finite set with n elements. Show that X has 2" distinct subsets. This exercise will, in effect, generalize the result of Exercise (ii). (a) Show that x"-1 is divisible by x - 1 for all n e N and where x1 is a natural number. (b) Find a formula for and use induction to show it holds for all n. (xiii) Consider the statement 'A (n): 2"< 2"-1. Prove that A(n) is false for all n e N Show that the inductive step holds, i.e. that A(k) A(k+1). Note that this shows that we need the initial case to be true to show the statement holds for all n. This is like 'The Moon is made of cheese implies that the Moon is a tasty snack' from Chapter 7. That is, 'A(n) A (n+1)' can be true even if both A (n) and A(n+ 1) are false. (xiv) Prove the statement on negation of quantifiers from Chapter 11: To negate a statement of the form Qix102x2... Qnxn P(x, x2...X), where Qr is Vor 3 for 1 <isn, we do the following: (a) Change every V to 3 and every 3 into V. (b) Replace P by its negation. xy)Show that al.- (This is the generalized Triangle Incquality.) (XvI) Suppose x1, x2, .. are non-negative real numbers. Prove that + Xn

Answer 1

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  divisible by & when nis (vii) to an odd show that n is natural number, Let le n = 1,3,5, 7, 9,.... m= 4 kt 1 and 4k+3 ; k = 0,1,2,... Let m= 4k+1 » n² = (4k+1} = 168²71-8k n²-1 = (16k²+1-8k) -1 = 16k? - Sk = 8k (2k-1) - ient is a multiple of 8 - 8 (-1) in 8 vdivides (7²1). Now, Il M= 4k + 3 & ² (4k+ 3) = 168279 +24k m² 1= (16k² + 9 + 24k) - = 168² +246 +8 m² = 8/2R²+3K+1) Again ²1 is a multiple of 8 → 81(021) Thus, when n is an odd natural number, 81(-1).
  
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  (xi) Let & be a finite set with an elements. To prove x has distinct a subsets. Let &- $a, ...., an} Empty set valso is a subsit of x. The subsets can be singletons ; sets with two elements, sets with three elements, and so on upto set with an elements (x itself). We can choose a object out of n objects in a ways Number of singletons = 6 Sets with two elements mean we have to choose 2 elements out of giren a elements. & Subsets, with a elements - Me Similarly, subsits with 3 elements = na and so on. subset with elements - Mn = 11x Eself) Remember Empty set is also a subset = "co= 1 Total subsets = m + " + + -- then = (2) using Binomial theorem for (itan for x=1 (1+x) = 6 + x + 2 + ... + (put x-1) we get an subsets
  * ⁿC₀ means choosing no element out of n which itself is a way of choosing!
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  (xii) (a) To prove (2-1) divides (221) & Neh and #1 EN We use proof by Induction. for Mal. It is same that (x-1) divides (x-1). Let us assume that for n=k ie. (x-1) divides the result holds (R1) (*) Now, to ie. prove for n=kti To prove (2-1) ktl divides ( 2 1) kt x-1 = kt k k x - x + x -1 b = mk (2-1) +121) from (*) ; we know (x-1) divides. (x2-1) (sk) = m(x-1) for some integerm → (41) +m14-) - (n) [a *m] (211) is a multiple of (x-1) I DEN REN & Am E ZI was mez) (2-1) divides del 'Hence, proved. kt
  
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  (6) method zo Let us use Long division divide (21) by (x-1). I w2 ta a t - + x + x + 1 2 w hy -1 n- 1-2 - 2 We continue the process till u- n- (1-2) th place ve when n=2 १२ x-1 저 ** () (*** ******) nt n-2 ta t - - - V -1 =
  

Now , we use induction to prove that it's correct.

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  n=1 LHS To prove 6-1) = (x-1) (1+x+r+ – – – + 2 ) I OR ( 1) = (x-1) (1+ x + 2? + - -- +2M) for - LHS - (2-1) = (x-1) (x+1) RHS = (x-1) (+2) = (x-1)(x+1) Thus, result holds for n=1 Let us rassume it is true for n=k (x - 1) = (x-1) (1+ x + x² + – – + x) ––– (*) kt Now, to prove that it holds for n=k+1) ktl kt2 il to ) prove d - 1 = (x-1) (+&+2²+ -- th kt2 kt2 kt kt Now LKS X 1 = x - x + 1 usung (*) Lis: **' (-) + (9-) (1+x+*+---+x)
  
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  LHS atit- - - t. (81) [ **' +41+2+3+ +-+7)] = (x-1) [1+x+2²+ - -- +2h + k +1] = RHS Hence proved.
  
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  (xv) to prove I bel = 1 x) using We prove it induction on n. d +---+ an To prove latay+ - - – tanl € 1lt Proof for n=1:- LHS - 1011 RHS = 18) and 10l =181 LHS = RHS for N=1. for n=2.- LHS - 18, +221 and RHS = 1 alt pal We know latb? = (ath? - a² +6+zab C a + b +2 191.16l - 1a²+ 1b²+ 21a1.16) = (191+ 16)² latbla (191+161) larbl = 191+1bl ( xray? nay only if & and y are non-negative) Here both latb) and (191+lbl) are always positive
  
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  121+22) s 12,1+1221 Proved for 1=2. Let ush rassume the result to be true for n=k. le 11, +42 +--- Tarla 19,1+122/+--+ lak) -- (*) Now, to prove result for M=(k+1) To prove prone - 19,+dat - – – + k + Akril a lat Hilt-- tortil LHS: 1 (0+2+ - - - tak tak til alaltazt--thklt/artil abuady proved triangle inequality for two elements assume (dit -. - + Uk) vas a single number Now usung 12+d2+ – – – +AR + lak til aldiltla21+ -- + hlk/+Hr+1) =R.HS. Hence proved 1 + ---- Hkl a lailt--- + laktil proved the Generalised triangle inequality using induction &n.
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