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Prove that if {a(n)} is a sequence of real numbers and a(n) --->a as n--->∞, then √a(n)---> √a as n --->∞.
Please help!
I am trying to find the cycle decompositions of some permutations. If you have 2 cyclic groups and you need to find the cycle decomposition of given permutations, I think I got it when there is only one permutations. You just map the numbers together until cycles form. But, when you have to join 2 groups together like a composition, all I know is let's say you have 2 cyclic groups you have to do permutations with (let's call them a and b). a and b individually will just be the numbers mapping to each other until you come back to the beginning number. That is one cycle, then you start with the next disjoint smallest number in the group of numbers.
Question: What if the permutation is a^2, ba, ab, ab^2, or a^b? I think applying the right side first is how you start, but I am really lost on this one. Please Help!
The class I am taking is Abstract Algebra (grad level). The problem I have having is with Basic Axioms and Groups. The problem states: Let G = {a+b√2 is an element of R|a,b are elements ofQ}. Prove that G is a group under addition. Do I pick random rational numbers to represent a and b to prove this theorem? I know you have to prove the identity and inverse are unique; the inverse of the inverse is the original element, and the binary operation (in this case, addition) is true under associative and commutative.
Another question; I am given the following info: Find the orders of each element Z/12Z. Now, all I know about this problem is n=12 and the class set is 0 bar through 11 bar. Should I add 12 to each class in order to find the remainder. I know the order is exponent on the number when the remainder is one.
I hope you can help me. This is time sensitive!
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