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anonymous

  • one year ago

Show that \[[x]+[x+\frac{1}{2}]=[2x]\] where \[[x]\] is the greatest integer of x

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  1. perl
    • one year ago
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    You can consider cases. When x is an integer, when x is not an integer

  2. anonymous
    • one year ago
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    @perl Would you please help me out

  3. Empty
    • one year ago
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    Here's a way you can solve this that I just came up with so I don't know if there's an easier way. First let's define \(0 \le d<\frac{1}{2}\) Then this means we can reach all numbers as either: \(x=n+d\) or \(x=n+d+\frac{1}{2}\). Why these weird choices? Cause when we plug them in we know no matter what \(d\) is we have: \[[n+d] = n\] \[[n+d+\frac{1}{2}] = n+1\] So we have two separate cases, plugging 'em in to: \[[x]+[x+\frac{1}{2}]=[2x]\] I think that will end up working?

  4. amoodarya
    • one year ago
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    "empty' give the answer i give another way you can also take two case below and check both of them \[\lfloor x \rfloor =2n\\ \lfloor x \rfloor=2n+1\]

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