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Could someone explain in detail how to prove this? show that: ceiling(lg(x+1)) = floor(lgx)+1 for integers x>=1 lgx means log 2 x I tried using
Could someone explain in detail how to prove this? show that: ceiling(lg(x+1)) = floor(lgx)+1 for integers x>=1
lgx means log2 x
I tried using let k = floor(lgx) then k <= lgx < k+1 therefore 2k <= x < 2k+1 (if we remove lg)
since x is an integer then: 2k < x + 1 <= 2k+1
therefore: k < lg(x+1) <= k+1 ,k = floor(lgx)
floor(lgx) < lg(x+1) <= floor(lgx)+1
I'm stuck here and I don't know if it makes sense
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