Recent content by rtw528

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    Abstract Algebra Proof (Cyclic cycles & order)

    Prove that if G is a group and aεG, then o(a-1)=o(a) This is all I have so far: Assume G is a group and aεG. Because G is a group a has an inverse in the group, a-1 s.t. aa-1=e, which is also in G. <a>={an|nεZ}. |<a>| is the number of elements in <a> before it cycles back. Basically all I've...
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    How to Choose Eigenvectors for Diagonalization

    This isn't really a homework question, but it is relevant to heling me finish my homework. When you are diagonalizing a matrix, how do you know what order to put the eigenvectors in. One of my homework problems is with the eigenvalues 1, 2, and 4. [-1] [1] is the matrix corresponding to the...
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    Proving a[n]≤2^n using Mathematical Induction

    Thanks so much for your help.
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    Proving a[n]≤2^n using Mathematical Induction

    And since 7<23, they are not equal but since the first two are, you use ≤ instead of = or <. Also 2k-223≥7(2k-2)
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    Proving a[n]≤2^n using Mathematical Induction

    2k-2(22+2+1), which is 2k-2(7). since neither this nor the other can get me to what I want, I made a mistake somewhere
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    Proving a[n]≤2^n using Mathematical Induction

    that a[k]+a[k-1]+a[k-2]≤2^k+2^(k-1)+2^(k-2). but how would I get to 2^(k+1) from here?
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    Proving a[n]≤2^n using Mathematical Induction

    1. Define a sequence of numbers in the following way: a[k]=a[k-1]+a[k-2]+a[k-3] for k≥3, s.t. a[0]=1, a[1]=2, a[2]=3, a[3]=6... *The numbers in brackets will be subscripts for the whole problem Prove that a[n]≤2^n using complete mathematical induction. This is what I have so far. We'll use...
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