y= ⌊ x^( 2) ⌋ + ⌊ x ⌋ ⇒ R


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Question Number 193971 by mnjuly1970 last updated on 24/Jun/23

$y = ⌊ x^{2} ⌋ + ⌊ x ⌋$ $\Rightarrow R_{y} = ? R a n g e$
	Answered by MM42 last updated on 24/Jun/23
	$R_y ={−1,0,1,2,3,...}$
	$R_{y} = {- 1, 0, 1, 2, 3, . . .}$
	Answered by mahdipoor last updated on 24/Jun/23
	$get x=(√(n^2 +m)) n∈W,0≤m∈R<2n+1 [x^2 ]+[x]=n^2 +[m]+n 0≤[m]<2n+1 R_(y1) ={n^2 +n+m∣n,m∈W,m≤2n} , get x=−(√(n^2 +m))≠0 n∈W,0≤m∈R<2n+1 [x^2 ]+[x]=n^2 +[m]−n−1 0≤[m]<2n+1 R_(y2) ={n^2 −n−1+m∣n,m∈W,m≤2n} , ∩⇒R_y ={r∈Z∣n^2 −n−1≤r≤n^2 +3n,n∈W} for ∀n∈W n^2 +3n≥(n+1)^2 −(n+1)−1 ⇒⇒R_y ={−1,0,1,2,...,∞}$
	$g e t x = \sqrt{n^{2} + m} n \in W, 0 ⩽ m \in R < 2 n + 1$ $[x^{2}] + [x] = n^{2} + [m] + n 0 ⩽ [m] < 2 n + 1$ $R_{y 1} = {n^{2} + n + m ∣ n, m \in W, m ⩽ 2 n}$ $,$ $g e t x = - \sqrt{n^{2} + m} \neq 0 n \in W, 0 ⩽ m \in R < 2 n + 1$ $[x^{2}] + [x] = n^{2} + [m] - n - 1 0 ⩽ [m] < 2 n + 1$ $R_{y 2} = {n^{2} - n - 1 + m ∣ n, m \in W, m ⩽ 2 n}$ $,$ $\cap \Rightarrow R_{y} = {r \in Z ∣ n^{2} - n - 1 ⩽ r ⩽ n^{2} + 3 n, n \in W}$ $f o r \forall n \in W n^{2} + 3 n ⩾ {(n + 1)}^{2} - (n + 1) - 1$ $\Rightarrow\Rightarrow R_{y} = {- 1, 0, 1, 2, . . ., \infty}$
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