Prove that: ⌊√n+1√n+√n+2⌋=⌊√n⌋, for all n∈N.
My solution:
Step 1:
First, note that the expression inside the floor function on the left can be rewritten such that we have:
√n+1√n+√n+2
=√n+1√n+√n+2⋅√n−√n+2√n−√n+2
=√n+√n−√n+2n−(n+2)
=√n−√n−√n+22
=√n+√n+2−√n2
=√n+√n+22
Step 2:
Next, note that for all n∈N, 4√n>−2 is always true. Algebraically manipulating it such that we get:
n+4√n+4>−2+n+4
(√n+2)2>n+2
√n+2>√n+2
Therefore we have:
√n+√n+2>√n+√n+2, which is
2√n+2>√n+√n+2
Step 3:
Note that we can set:
√n+√n<√n+√n+2<2√n+2
which is
2√n<√n+√n+2<2(√n+1)
√n<√n+√n+22<√n+1
Thus, we can conclude at this juncture that ⌊√n+√n+22⌋=⌊√n+1√n+√n+2⌋=⌊√n⌋ and we're hence done.
i dont Think that you can draw that conclusion simple because sqrt(x) is not necessarily an integer. för example if x € (2.3 , 3.3) then the integer part of x could either 2 or 3.
ReplyDeleteSébastien