How can one prove that if x is an integer greater than 2, then x/(x-1) is a not an integer?
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How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?
Intuitively I can see this is true but how to prove it?
elementary-number-theory
asked Nov 27 at 8:31
pirsquare
379319
add a comment |
up vote
0
down vote
favorite
How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?
Intuitively I can see this is true but how to prove it?
elementary-number-theory
asked Nov 27 at 8:31
pirsquare
379319
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?
Intuitively I can see this is true but how to prove it?
elementary-number-theory
asked Nov 27 at 8:31
pirsquare
379319
How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?
Intuitively I can see this is true but how to prove it?
elementary-number-theory
elementary-number-theory
asked Nov 27 at 8:31
pirsquare
379319
asked Nov 27 at 8:31
pirsquare
379319
edited Nov 27 at 8:40
asked Nov 27 at 8:31
pirsquare
379319
asked Nov 27 at 8:31
pirsquare
379319
asked Nov 27 at 8:31
pirsquare
379319
379319
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
add a comment |
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
2
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
add a comment |
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
222k15155273
add a comment |
up vote
2
down vote
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.1k981179
add a comment |
up vote
2
down vote
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
18.8k2736
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
222k15155273
add a comment |
up vote
5
down vote
accepted
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
222k15155273
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
222k15155273
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
222k15155273
answered Nov 27 at 8:34
lab bhattacharjee
222k15155273
answered Nov 27 at 8:34
lab bhattacharjee
222k15155273
answered Nov 27 at 8:34
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
up vote
2
down vote
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.1k981179
add a comment |
up vote
2
down vote
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.1k981179
add a comment |
up vote
2
down vote
up vote
2
down vote
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.1k981179
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.1k981179
answered Nov 27 at 8:47
Mark Bennet
80.1k981179
answered Nov 27 at 8:47
Mark Bennet
80.1k981179
answered Nov 27 at 8:47
Mark Bennet
80.1k981179
80.1k981179
add a comment |
add a comment |
up vote
2
down vote
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
18.8k2736
add a comment |
up vote
2
down vote
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
18.8k2736
add a comment |
up vote
2
down vote
up vote
2
down vote
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
18.8k2736
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
18.8k2736
answered Nov 27 at 9:06
farruhota
18.8k2736
answered Nov 27 at 9:06
farruhota
18.8k2736
answered Nov 27 at 9:06
farruhota
18.8k2736
18.8k2736
add a comment |
add a comment |
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2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39