Recall that: ged: NN → N gcd(a,0) = a. gcd(a,b) = gcd(b, mod(a,b)), if b >...
1. Recall the following theorem. Theorem 1. Let a, b, m,n e N, m, n > 0 and ged(m,n) = 1. There erists a unique r e Zmn such that the following holds. x = a (mod m) x = b (mod n) please show that such solution is unique.
2,3,4,5,6 please 2. Use the Euclidean algorithm to find the following: a gcd(100, 101) b. ged(2482, 7633) 3. Prove that if a = bq+r, then ged(a, b) = ged(b,r). such that sa tb ged(a,b) for the following pairs 4. Use Bézout's theorem to find 8 and a. 33, 44 b. 101, 203 c. 10001, 13422 5. Prove by induction that if p is prime and plaja... An, then pla, for at least one Q. (Hint: use n = 2 as...
Recall from class that the Fibonacci numbers are defined as follows: fo = 0,fi-1 and for all n fn-n-1+fn-2- 2, (a) Let nEN,n 24. Prove that when we divide In by f-1, the quotient is 1 and the remainder is fn-2 (b) Prove by induction/recursion that the Euclidean Algorithm takes n-2 iterations to calculate gcd(fn,fn-1) for n 2 3. Check your answer for Question 1 against this. Recall from class that the Fibonacci numbers are defined as follows: fo =...
Problem 44) Prove: n!> 2" for n24. Problem 45) Prove by induction: For n>0·AT- i=1
.n= n(n-1)(n+1) for all n > 2. 12. Use induction to prove (1 : 2) +(2-3)+(3-4) +...+(n-1).n [9 points) 3
We know that we can reduce the base of an exponent modulo m: a(a mod m)k (mod m). But the same is not true of the exponent itself! That is, we cannot write aa mod m (mod m). This is easily seen to be false in general. Consider, for instance, that 210 mod 3 1 but 210 mod 3 mod 3 21 mod 3-2. The correct law for the exponent is more subtle. We will prove it in steps (a)...
1 For each of the following pairs of numbers a and b, calculate and find integers r and s such ged (a; b) by Eucledian algorithm that gcd(a; b) = ra + sb. ia= 203, b-91 ii a = 21, b=8 2 Prove that for n 2 1,2+2+2+2* +...+2 -2n+1 -2 3 Prove that Vn 2 1,8" -3 is divisible by 5. 4 Prove that + n(n+1) = nnīYn E N where N is the set of all positive integers....
Use induction to prove that 0–0 4j3 = n4 + 2n3 + n2 where n > 0.
Use induction and Pascal's identity to prove that (7) = 2" where n > 0.
| Prove that for n e N, n > 0, we have 1 x 1!+ 2 x 2!+... tnx n! = (n + 1)! - 1.