Easy but confusing
Find the solution set to ⌊ 2 x − 1 ⌋ < 5
Note:
⌊
x
⌋
represents greatest integer not exceeding x.
x
∈
R
(
v
,
h
)
is an interval representing all real numbers between v and h excluding v and h.
[
v
,
h
]
is an interval representing all real numbers between v and h including v and h.
(
v
,
h
]
is an interval representing all real numbers between v and h excluding v .
[
v
,
h
)
is an interval representing all real numbers between v and h excluding h.
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2 solutions
Lemma proof brief explanation of yours is nice !
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The lemma was too trivial to state though.
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Nope, the lemma may seem trivial but a rigorous proof is completely non-trivial. If possible, state the proof please bro.
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@Venkata Karthik Bandaru – I have added the proof in the comments. Go check it out! :)
If anybody wants me to present a full proof of that lemma, reply to this comment. I'll add the proof in the comments then.
Proof:
Let y ∈ R such that y = x + δ , δ ∈ [ 0 , 1 ) and x = ⌊ y ⌋ . Then, depending upon the value of y , we have,
⌊ y ⌋ < a ⟹ x = a − γ , γ ∈ N ⟹ x + δ = a − γ + δ
∵ δ ∈ [ 0 , 1 ) ⟹ a − γ + δ ∈ [ a − γ , a − γ + 1 ) ⟹ x + δ ∈ [ a − γ , a − γ + 1 ) ⟹ x + δ < a + 1 − γ
We have γ ∈ N and its value depends upon the value of y . The strict inequality with the supremum (upper bound) for y = x + δ is obviously attained when γ = γ min = 1 . Thus, we obtain the inequality with the upper bound as,
x + δ < a + 1 − 1 ⟹ x + δ < a ⟹ y < a
Hence, the lemma is established and proved.
Just wanted to share some immediate consequences of the lemma proved by Prasun....
Please correct if you find any mistake and upvote if you like !
Lemma: If y ∈ R and ⌊ y ⌋ < a where a ∈ Z + , then we can remove the floor function and the inequality will still hold, i.e., y < a
Proof of Lemma: The lemma can be proven trivially using the definition of the floor function. A more formal and rigorous proof of this trivial lemma can be presented by making the substitution y = x + δ , δ ∈ [ 0 , 1 ) and then introducing a value γ ∈ N when the floor is removed from the inequality and ultimately concluding that y = x + δ < a . A rigorous proof of this is given in the comments.
Using the lemma on the given data, we have,
⌊ 2 x − 1 ⌋ < 5 ⟹ 2 x − 1 < 5 ⟹ x < 3 ⟹ x ∈ ( − ∞ , 3 )