Question 1. Let y be a positive number. Prove that ⌈ ⌈y⌉
⌉ = ⌈ y2 ⌉.
2
Proof. If y ∈ Z, then readily y = ⌈y⌉ and the claim of the question follows.
If y ∈
/ Z, then
⌈y⌉ − 1 < y < ⌈y⌉.
(1)
Consider two cases: even ⌈y⌉ and odd ⌈y⌉.
⌉ = ⌈ ⌈y⌉
⌉ and thus we obtain from (1) that
When ⌈y⌉ is even, ⌈ ⌈y⌉−1
2
2
⌈
⌈y⌉
y
⌈y⌉
⌈y⌉ − 1
⌉=⌈
⌉≤⌈ ⌉≤⌈
⌉.
2
2
2
2
⌉ = ⌈ y2 ⌉ when ⌈y⌉ is even.
Consequently, ⌈ ⌈y⌉
2
Now when ⌈y⌉ is odd, ⌈ ⌈y⌉−1
⌉ = ⌈ ⌈y⌉
⌉ − 1 < ⌈ y2 ⌉ and thus we obtain from
2
2
(1) that
⌈y⌉ − 1
⌈y⌉
y
⌈y⌉
⌈
⌉=⌈
⌉−1<⌈ ⌉≤⌈
⌉.
(2)
2
2
2
2
⌉ = ⌈ y2 ⌉ when ⌈y⌉ is odd.
Since ⌈ y2 ⌉ is an integer, (2) implies that ⌈ ⌈y⌉
2
Question 2. Let d be a positive integer and define ⌈ d2 ⌉(i) recursively as
follows: ⌈ d2 ⌉(1) = ⌈ d2 ⌉ and for i ∈ Z+ \ {1}, ⌈ d2 ⌉(i) = ⌈
⌈ d2 ⌉(i) = ⌈ 2di ⌉ for i ∈ Z+ .
⌈ d2 ⌉(i−1)
⌉.
2
Prove that
Proof. Let us prove the claim of the question inductively.
Basis. By the definition, ⌈ d2 ⌉(1) = ⌈ d2 ⌉.
Hypothesis. We assume that ⌈ d2 ⌉(i) = ⌈ 2di ⌉ and turn to prove that ⌈ d2 ⌉(i+1) =
d
⌉.
⌈ 2i+1
By the definition, ⌈ d2 ⌉(i+1) = ⌈
⌈ d2 ⌉(i)
⌉,
2
hence by the induction hypothesis,
d (i+1) l ⌈ 2di ⌉ m
.
⌈ ⌉
=
2
2
l⌈ d ⌉m l
m
l
m
d
d
By Question 1, 22i = 2i+1
, thus ⌈ d2 ⌉(i+1) = 2i+1
, which concludes the
proof.
1