a) Consider the map P: N-->N which sends (a,b) to (2^a)(3^b). Prove that P is in

Question

a) Consider the map P: N-->N which sends (a,b) to (2^a)(3^b). Prove that P is injective. Conclude that NxN is countably infinite. (*Note: N is the natural numbers)

b) For all n let Jn = [(n-1)/n, (n+1)/n]. Prove that ?Jn = {1} (*Note: ?Jn is from n=1 to +infinity)

c) Let Xn be a sequence satisfying that if m does not equal n "within" N are any distinct natural numbers, then

1/(2*m*n) "is less than or equal to" absolute value (Xm-Xn) "is less than or equal to" 1/(m+n). Show that (Xn) is convergent but not contractive.

Explanation / Answer

To show that the map is injective, let P(a,b) = P(c,d). Then we need to show (a,b)=(c,d). Since P(a,b) = P(c,d) then 2^a 3^b = 2^c 3^d, so dividing both sides we get, 2^(a-c) 3^(b-d) = 1. Since 2 and 3 are primes, then the product of 2^(a-c) 3^(b-d) can only be 1 if 2^(a-c) = 1 and 3^(b-d)=1. So a-c = 0 and b-d = 0, hence a=c and b=d. Finally, this shows (a,b)=(c,d).

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