Mixing
Prove that ${1over x_1}+{1over x_2}+dots+{1over x_n}lt3$ if no $x_j=10^kx_i+n$ where $x_i,k,ninmathbb{Z^+}$...
-1
$begingroup$
Prove that ${1over x_1}+{1over x_2}+dots+{1over x_n}lt3$ if no $x_j=10^kx_i+n$ where $x_i,k,ninmathbb{Z^+}$, $n<10$, and $x_inot=x_j$
I have attempted this question multiple times and have barely reached anything. I tried to assume WLOG that $x_1le x_2ledotsle x_n$ however I could not continue. I am still new to such inequality questions so any help would be appreciated. Thank you anyways.
algebra-precalculus inequality
asked Jan 29 at 14:45
user587054user587054
57911
$endgroup$
add a comment |
-1
$begingroup$
Prove that ${1over x_1}+{1over x_2}+dots+{1over x_n}lt3$ if no $x_j=10^kx_i+n$ where $x_i,k,ninmathbb{Z^+}$, $n<10$, and $x_inot=x_j$
I have attempted this question multiple times and have barely reached anything. I tried to assume WLOG that $x_1le x_2ledotsle x_n$ however I could not continue. I am still new to such inequality questions so any help would be appreciated. Thank you anyways.
algebra-precalculus inequality
asked Jan 29 at 14:45
user587054user587054
57911
$endgroup$
2
$begingroup$
Suppose $n=4$ and $x_1=x_2=x_3=x_4=1$. Then it is certainly true that no $x_j=10^kx_i+4$ but $sum frac 1{x_i}=4$. Or am I misreading?
$endgroup$
– lulu
Jan 29 at 15:00
$begingroup$
I forgot to add that they are distinct
$endgroup$
– user587054
Jan 29 at 15:02
1
$begingroup$
And are they meant to be positive integers? Please edit your post to include all the conditions you have in mind.
$endgroup$
– lulu
Jan 29 at 15:03
1
$begingroup$
Ok. Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then again it is clearly true that no $x_j=10^kx_i+11$ and $sum frac 1{x_i}=3.019877345>3$.
$endgroup$
– lulu
Jan 29 at 15:11
add a comment |
-1
-1
-1
$begingroup$
Prove that ${1over x_1}+{1over x_2}+dots+{1over x_n}lt3$ if no $x_j=10^kx_i+n$ where $x_i,k,ninmathbb{Z^+}$, $n<10$, and $x_inot=x_j$
I have attempted this question multiple times and have barely reached anything. I tried to assume WLOG that $x_1le x_2ledotsle x_n$ however I could not continue. I am still new to such inequality questions so any help would be appreciated. Thank you anyways.
algebra-precalculus inequality
asked Jan 29 at 14:45
user587054user587054
57911
$endgroup$
Prove that ${1over x_1}+{1over x_2}+dots+{1over x_n}lt3$ if no $x_j=10^kx_i+n$ where $x_i,k,ninmathbb{Z^+}$, $n<10$, and $x_inot=x_j$
I have attempted this question multiple times and have barely reached anything. I tried to assume WLOG that $x_1le x_2ledotsle x_n$ however I could not continue. I am still new to such inequality questions so any help would be appreciated. Thank you anyways.
algebra-precalculus inequality
algebra-precalculus inequality
asked Jan 29 at 14:45
user587054user587054
57911
asked Jan 29 at 14:45
user587054user587054
57911
edited Jan 29 at 15:16
user587054
asked Jan 29 at 14:45
user587054user587054
57911
asked Jan 29 at 14:45
user587054user587054
57911
asked Jan 29 at 14:45
user587054user587054
57911
57911
2
$begingroup$
Suppose $n=4$ and $x_1=x_2=x_3=x_4=1$. Then it is certainly true that no $x_j=10^kx_i+4$ but $sum frac 1{x_i}=4$. Or am I misreading?
$endgroup$
– lulu
Jan 29 at 15:00
$begingroup$
I forgot to add that they are distinct
$endgroup$
– user587054
Jan 29 at 15:02
1
$begingroup$
And are they meant to be positive integers? Please edit your post to include all the conditions you have in mind.
$endgroup$
– lulu
Jan 29 at 15:03
1
$begingroup$
Ok. Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then again it is clearly true that no $x_j=10^kx_i+11$ and $sum frac 1{x_i}=3.019877345>3$.
$endgroup$
– lulu
Jan 29 at 15:11
add a comment |
2
$begingroup$
Suppose $n=4$ and $x_1=x_2=x_3=x_4=1$. Then it is certainly true that no $x_j=10^kx_i+4$ but $sum frac 1{x_i}=4$. Or am I misreading?
$endgroup$
– lulu
Jan 29 at 15:00
$begingroup$
I forgot to add that they are distinct
$endgroup$
– user587054
Jan 29 at 15:02
1
$begingroup$
And are they meant to be positive integers? Please edit your post to include all the conditions you have in mind.
$endgroup$
– lulu
Jan 29 at 15:03
1
$begingroup$
Ok. Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then again it is clearly true that no $x_j=10^kx_i+11$ and $sum frac 1{x_i}=3.019877345>3$.
$endgroup$
– lulu
Jan 29 at 15:11
2
2
$begingroup$
Suppose $n=4$ and $x_1=x_2=x_3=x_4=1$. Then it is certainly true that no $x_j=10^kx_i+4$ but $sum frac 1{x_i}=4$. Or am I misreading?
$endgroup$
– lulu
Jan 29 at 15:00
$begingroup$
Suppose $n=4$ and $x_1=x_2=x_3=x_4=1$. Then it is certainly true that no $x_j=10^kx_i+4$ but $sum frac 1{x_i}=4$. Or am I misreading?
$endgroup$
– lulu
Jan 29 at 15:00
$begingroup$
I forgot to add that they are distinct
$endgroup$
– user587054
Jan 29 at 15:02
$begingroup$
I forgot to add that they are distinct
$endgroup$
– user587054
Jan 29 at 15:02
1
1
$begingroup$
And are they meant to be positive integers? Please edit your post to include all the conditions you have in mind.
$endgroup$
– lulu
Jan 29 at 15:03
$begingroup$
And are they meant to be positive integers? Please edit your post to include all the conditions you have in mind.
$endgroup$
– lulu
Jan 29 at 15:03
1
1
$begingroup$
Ok. Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then again it is clearly true that no $x_j=10^kx_i+11$ and $sum frac 1{x_i}=3.019877345>3$.
$endgroup$
– lulu
Jan 29 at 15:11
$begingroup$
Ok. Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then again it is clearly true that no $x_j=10^kx_i+11$ and $sum frac 1{x_i}=3.019877345>3$.
$endgroup$
– lulu
Jan 29 at 15:11
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
As stated this is false:
Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then it is clearly true that no $x_j=10^kx_i+11$ (for $kin mathbb Z^+$) and $$sum_{i=1}^{11} frac 1{i}=3.019877345>3$$
answered Jan 29 at 15:12
lulululu
43.3k25080
$endgroup$
$begingroup$
forgot to write that $n<10$
$endgroup$
– user587054
Jan 29 at 15:17
1
$begingroup$
It's hard to answer questions when the terms keep changing. This new rule can't be right, since $sum_{i=1}^{9}frac 1i=2.828968254<3$.
$endgroup$
– lulu
Jan 29 at 15:18
$begingroup$
im sorry, hopefully now all the terms are right.
$endgroup$
– user587054
Jan 29 at 15:20
1
$begingroup$
As I say, with $n<10$ the problem is trivial, since $sum_{i=1}^9frac 19<3$.
$endgroup$
– lulu
Jan 29 at 15:21
2
$begingroup$
At this point, I'm guessing that you incorrectly used the variable $n$ to mean two different things...you use it as the number of the $x_i$ and you use it as the "remainder" in the $x_j=10^kx_i+n$ expression.
$endgroup$
– lulu
Jan 29 at 15:23
add a comment |
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1 Answer
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1 Answer
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1
$begingroup$
As stated this is false:
Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then it is clearly true that no $x_j=10^kx_i+11$ (for $kin mathbb Z^+$) and $$sum_{i=1}^{11} frac 1{i}=3.019877345>3$$
answered Jan 29 at 15:12
lulululu
43.3k25080
$endgroup$
$begingroup$
forgot to write that $n<10$
$endgroup$
– user587054
Jan 29 at 15:17
1
$begingroup$
It's hard to answer questions when the terms keep changing. This new rule can't be right, since $sum_{i=1}^{9}frac 1i=2.828968254<3$.
$endgroup$
– lulu
Jan 29 at 15:18
$begingroup$
im sorry, hopefully now all the terms are right.
$endgroup$
– user587054
Jan 29 at 15:20
1
$begingroup$
As I say, with $n<10$ the problem is trivial, since $sum_{i=1}^9frac 19<3$.
$endgroup$
– lulu
Jan 29 at 15:21
2
$begingroup$
At this point, I'm guessing that you incorrectly used the variable $n$ to mean two different things...you use it as the number of the $x_i$ and you use it as the "remainder" in the $x_j=10^kx_i+n$ expression.
$endgroup$
– lulu
Jan 29 at 15:23
add a comment |
1
$begingroup$
As stated this is false:
Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then it is clearly true that no $x_j=10^kx_i+11$ (for $kin mathbb Z^+$) and $$sum_{i=1}^{11} frac 1{i}=3.019877345>3$$
answered Jan 29 at 15:12
lulululu
43.3k25080
$endgroup$
$begingroup$
forgot to write that $n<10$
$endgroup$
– user587054
Jan 29 at 15:17
1
$begingroup$
It's hard to answer questions when the terms keep changing. This new rule can't be right, since $sum_{i=1}^{9}frac 1i=2.828968254<3$.
$endgroup$
– lulu
Jan 29 at 15:18
$begingroup$
im sorry, hopefully now all the terms are right.
$endgroup$
– user587054
Jan 29 at 15:20
1
$begingroup$
As I say, with $n<10$ the problem is trivial, since $sum_{i=1}^9frac 19<3$.
$endgroup$
– lulu
Jan 29 at 15:21
2
$begingroup$
At this point, I'm guessing that you incorrectly used the variable $n$ to mean two different things...you use it as the number of the $x_i$ and you use it as the "remainder" in the $x_j=10^kx_i+n$ expression.
$endgroup$
– lulu
Jan 29 at 15:23
add a comment |
1
1
1
$begingroup$
As stated this is false:
Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then it is clearly true that no $x_j=10^kx_i+11$ (for $kin mathbb Z^+$) and $$sum_{i=1}^{11} frac 1{i}=3.019877345>3$$
answered Jan 29 at 15:12
lulululu
43.3k25080
$endgroup$
As stated this is false:
Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then it is clearly true that no $x_j=10^kx_i+11$ (for $kin mathbb Z^+$) and $$sum_{i=1}^{11} frac 1{i}=3.019877345>3$$
answered Jan 29 at 15:12
lulululu
43.3k25080
answered Jan 29 at 15:12
lulululu
43.3k25080
answered Jan 29 at 15:12
lulululu
43.3k25080
answered Jan 29 at 15:12
lulululu
43.3k25080
43.3k25080
$begingroup$
forgot to write that $n<10$
$endgroup$
– user587054
Jan 29 at 15:17
1
$begingroup$
It's hard to answer questions when the terms keep changing. This new rule can't be right, since $sum_{i=1}^{9}frac 1i=2.828968254<3$.
$endgroup$
– lulu
Jan 29 at 15:18
$begingroup$
im sorry, hopefully now all the terms are right.
$endgroup$
– user587054
Jan 29 at 15:20
1
$begingroup$
As I say, with $n<10$ the problem is trivial, since $sum_{i=1}^9frac 19<3$.
$endgroup$
– lulu
Jan 29 at 15:21
2
$begingroup$
At this point, I'm guessing that you incorrectly used the variable $n$ to mean two different things...you use it as the number of the $x_i$ and you use it as the "remainder" in the $x_j=10^kx_i+n$ expression.
$endgroup$
– lulu
Jan 29 at 15:23
add a comment |
$begingroup$
forgot to write that $n<10$
$endgroup$
– user587054
Jan 29 at 15:17
1
$begingroup$
It's hard to answer questions when the terms keep changing. This new rule can't be right, since $sum_{i=1}^{9}frac 1i=2.828968254<3$.
$endgroup$
– lulu
Jan 29 at 15:18
$begingroup$
im sorry, hopefully now all the terms are right.
$endgroup$
– user587054
Jan 29 at 15:20
1
$begingroup$
As I say, with $n<10$ the problem is trivial, since $sum_{i=1}^9frac 19<3$.
$endgroup$
– lulu
Jan 29 at 15:21
2
$begingroup$
At this point, I'm guessing that you incorrectly used the variable $n$ to mean two different things...you use it as the number of the $x_i$ and you use it as the "remainder" in the $x_j=10^kx_i+n$ expression.
$endgroup$
– lulu
Jan 29 at 15:23
$begingroup$
forgot to write that $n<10$
$endgroup$
– user587054
Jan 29 at 15:17
$begingroup$
forgot to write that $n<10$
$endgroup$
– user587054
Jan 29 at 15:17
1
1
$begingroup$
It's hard to answer questions when the terms keep changing. This new rule can't be right, since $sum_{i=1}^{9}frac 1i=2.828968254<3$.
$endgroup$
– lulu
Jan 29 at 15:18
$begingroup$
It's hard to answer questions when the terms keep changing. This new rule can't be right, since $sum_{i=1}^{9}frac 1i=2.828968254<3$.
$endgroup$
– lulu
Jan 29 at 15:18
$begingroup$
im sorry, hopefully now all the terms are right.
$endgroup$
– user587054
Jan 29 at 15:20
$begingroup$
im sorry, hopefully now all the terms are right.
$endgroup$
– user587054
Jan 29 at 15:20
1
1
$begingroup$
As I say, with $n<10$ the problem is trivial, since $sum_{i=1}^9frac 19<3$.
$endgroup$
– lulu
Jan 29 at 15:21
$begingroup$
As I say, with $n<10$ the problem is trivial, since $sum_{i=1}^9frac 19<3$.
$endgroup$
– lulu
Jan 29 at 15:21
2
2
$begingroup$
At this point, I'm guessing that you incorrectly used the variable $n$ to mean two different things...you use it as the number of the $x_i$ and you use it as the "remainder" in the $x_j=10^kx_i+n$ expression.
$endgroup$
– lulu
Jan 29 at 15:23
$begingroup$
At this point, I'm guessing that you incorrectly used the variable $n$ to mean two different things...you use it as the number of the $x_i$ and you use it as the "remainder" in the $x_j=10^kx_i+n$ expression.
$endgroup$
– lulu
Jan 29 at 15:23
add a comment |
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2
$begingroup$
Suppose $n=4$ and $x_1=x_2=x_3=x_4=1$. Then it is certainly true that no $x_j=10^kx_i+4$ but $sum frac 1{x_i}=4$. Or am I misreading?
$endgroup$
– lulu
Jan 29 at 15:00
$begingroup$
I forgot to add that they are distinct
$endgroup$
– user587054
Jan 29 at 15:02
1
$begingroup$
And are they meant to be positive integers? Please edit your post to include all the conditions you have in mind.
$endgroup$
– lulu
Jan 29 at 15:03
1
$begingroup$
Ok. Taking $n=11$ and $(x_1,cdots,x_n)=(1,2,3,cdots, 11)$ then again it is clearly true that no $x_j=10^kx_i+11$ and $sum frac 1{x_i}=3.019877345>3$.
$endgroup$
– lulu
Jan 29 at 15:11