Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists …
Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that 
for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
algebra-precalculus functions discrete-mathematics floor-function
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
64.8k8158246
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
add a comment |
Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
algebra-precalculus functions discrete-mathematics floor-function
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
64.8k8158246
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
– Andrés E. Caicedo
Dec 28 '18 at 14:07
add a comment |
Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
algebra-precalculus functions discrete-mathematics floor-function
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
64.8k8158246
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
algebra-precalculus functions discrete-mathematics floor-function
algebra-precalculus functions discrete-mathematics floor-function
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
64.8k8158246
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
64.8k8158246
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
64.8k8158246
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
64.8k8158246
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
64.8k8158246
64.8k8158246
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
464310
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
– Andrés E. Caicedo
Dec 28 '18 at 14:07
add a comment |
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
– Andrés E. Caicedo
Dec 28 '18 at 14:07
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
– Andrés E. Caicedo
Dec 28 '18 at 14:07
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
– Andrés E. Caicedo
Dec 28 '18 at 14:07
add a comment |
1 Answer
1
active
oldest
votes
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
968414
1
Can the downvoter please explain?
– Lucas Henrique
Dec 28 '18 at 4:58
add a comment |
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1 Answer
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Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
968414
1
Can the downvoter please explain?
– Lucas Henrique
Dec 28 '18 at 4:58
add a comment |
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
968414
1
Can the downvoter please explain?
– Lucas Henrique
Dec 28 '18 at 4:58
add a comment |
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
968414
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
968414
edited Dec 28 '18 at 13:55
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
968414
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
968414
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
968414
968414
1
Can the downvoter please explain?
– Lucas Henrique
Dec 28 '18 at 4:58
add a comment |
1
Can the downvoter please explain?
– Lucas Henrique
Dec 28 '18 at 4:58
1
1
Can the downvoter please explain?
– Lucas Henrique
Dec 28 '18 at 4:58
Can the downvoter please explain?
– Lucas Henrique
Dec 28 '18 at 4:58
add a comment |
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What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
– Andrés E. Caicedo
Dec 28 '18 at 14:07