Question : I want to prove that an infinite subset of a denumerable set is itself denumerable. So say the infinite subset is S, and the superset is A. Now I have to show there is a bijection between S and the natural numbers Z+.
. How do you show 1-1 and onto between S and Z+. Please explain in depth.
And please , explain what happens if S is not a subset of the natural numbers. For example S could be a subset of the set of algebraic numbers. Then we have a problem. THe proof that S is denumerable hing..
Answer : If you already know that every infinite subset of Z+ is denumerable, then you can use this fact to prove that every infinite subset of an arbitrary denumerable set is denumerable.
Using your notation, since A is denumerable, there exists a bijection f from A to Z+. Since S is an infinite subset of A, f(S) is an infinite subset of Z+. But every infinite subset of Z+ is denumerable, so there exists a bijection g from f(S) to Z+. Then, the composite function gf is a bijection from S to Z+, thus S is denumerable.
Question : Find the sum of the infinite geometric series, a(sub1) = -5, r =^1/6
-6? r= ^1/6 is the rate
Answer : An infinite geometric series has the solution of s=a/(1-r), where a is the initial value and r is the common ratio, which must be between -1 and 1, but not 1, 0, nor -1. So with your values, of a = -5 and r = 1/6,
s = -5/(1-1/6) = -5/(5/6) = -5 (6/5) = -6
BTW, what is with the '^' in your definition of r ?
Question : I have a package called 'Math Success' by Topics which covers Algebra and geometry but I don't think it is a good program. Any other recommendations?
Answer : The Algebra Helper software can help you with your homework. It makes your homework faster to do and easier to learn...
Calclus Software: http://www.brightideassoftware.com/