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Math Help - Proof Question

  1. #1
    Newbie Lazarath's Avatar
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    Proof Question

    There exists a c in R+ st. for all n in N, n>=2 --> 3n^2 - 4n >= cn^2
    How would I approach a proof idea, then find an appropriate value for c and then a formal well structured proof. I intuitively figured out c <= 2 && c > 1, But how to begin?

    Any suggestions or comments?

    Thanks
    Laz
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  2. #2
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    Quote Originally Posted by Lazarath
    There exists a c in R+ st. for all n in N, n>=2 --> 3n^2 - 4n >= cn^2
    How would I approach a proof idea, then find an appropriate value for c and then a formal well structured proof. I intuitively figured out c <= 2 && c > 1, But how to begin?

    Any suggestions or comments?

    Thanks
    Laz
    You need to find a c\in \mathbb{R}^+ such as, 3n^2-4n\geq cn^2 for all n\geq 2. Notice that,
    3n^2-4n\geq cn^2
    if and only if, (divide by n\geq 2) thus, postive the sign does not change
    3n-4\geq cn
    if and only if,
    3n-cn\geq 4
    if and only if,
    n(3-c)\geq 4
    if and only if, for all n\geq 2 \in \mathbb{Z}
    3-c\geq \frac{4}{n} (1)

    If a<b and (1) is true for a then it is true for b too. Thus, if (1) is true for the smallest n then it is true for all integers bigger than it. Thus, it must be true for 2.
    Thus, 3-c\geq \frac{4}{2}=2
    Thus, solving for c\leq 1.

    Now we check whether c\leq 1 is a solution. Thus, 3-c\leq 2=\frac{4}{2}. Since it is true for n=2 it is true for all n\geq 2 as explained in the previous paragraph.

    \mathbb{Q}.\mathbb{E}.\mathbb{D}.
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  3. #3
    Newbie Lazarath's Avatar
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    Whitby Ontario
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    Thank you

    Just wanted to say thanks for the timely reply. Never thought I'd be in University of Toronto at 40yrs trying to wrap my head around this!

    Thanks Again,

    Lazarath
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