- #1
- 8
- 0
So I've got a calculus test in a week, and I'm studying for it but I can't understand some examples our professor has given us. So, he says:
1) A = { x[itex]\in[/itex] ℝ: (x-a)(x-b)(x-c) < 0 } , a<b<c. Find the supA and infA.
In the solution of his example he says. It is easy to see that A = (-∞,a)[itex]\cup[/itex](b,c) so infA = -∞ and supA = c. How did he find that A = (-∞,a)[itex]\cup[/itex](b,c) ?
2) B = {1 +(-1)n:n[itex]\in[/itex] N}. Find the supB and infB.
In the solution he says. Obviously B={0,1} which we can compute if we make n=2k (even) and n=2k+1 (odd)
so infB=0 and supB=1. So if n=2k wouldn't B be 2? Is this a mistake my professor made?
1) A = { x[itex]\in[/itex] ℝ: (x-a)(x-b)(x-c) < 0 } , a<b<c. Find the supA and infA.
In the solution of his example he says. It is easy to see that A = (-∞,a)[itex]\cup[/itex](b,c) so infA = -∞ and supA = c. How did he find that A = (-∞,a)[itex]\cup[/itex](b,c) ?
2) B = {1 +(-1)n:n[itex]\in[/itex] N}. Find the supB and infB.
In the solution he says. Obviously B={0,1} which we can compute if we make n=2k (even) and n=2k+1 (odd)
so infB=0 and supB=1. So if n=2k wouldn't B be 2? Is this a mistake my professor made?