Convert any positive number to decimal $x$ such that $0lt x lt 1$.
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This is a very basic question, I understand that.
So I have these numbers: 20, 0.75, 1.25. If you sort them, it would look like this:
20
1.25
0.75
I need to convert any number that's greater than 1 to a decimal that's less than 1 and greater than 0, and still sort like above.
So I thought about dividing 1 by that number. But naturally, that's not going to work with the 0.75, so 0.75 will be greater than 20.
So what formula can I apply to any positive number so that they all are less than 1 and greater than 0, and they still sort correctly with this new value (i.e. #1: 20, #2: 1.25, #3: 0.75)
algebra-precalculus elementary-functions
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up vote
-1
down vote
favorite
This is a very basic question, I understand that.
So I have these numbers: 20, 0.75, 1.25. If you sort them, it would look like this:
20
1.25
0.75
I need to convert any number that's greater than 1 to a decimal that's less than 1 and greater than 0, and still sort like above.
So I thought about dividing 1 by that number. But naturally, that's not going to work with the 0.75, so 0.75 will be greater than 20.
So what formula can I apply to any positive number so that they all are less than 1 and greater than 0, and they still sort correctly with this new value (i.e. #1: 20, #2: 1.25, #3: 0.75)
algebra-precalculus elementary-functions
Is there a minimum and maximum value that your numbers can take?
â Jared Goguen
Aug 6 at 18:01
3
Part of the question says "any number that's greater than 1," and another part says "apply to any number." You should probably reconcile these two different criteria, especially since the chosen answer fulfills one of them and not the other.
â John
Aug 6 at 18:03
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
This is a very basic question, I understand that.
So I have these numbers: 20, 0.75, 1.25. If you sort them, it would look like this:
20
1.25
0.75
I need to convert any number that's greater than 1 to a decimal that's less than 1 and greater than 0, and still sort like above.
So I thought about dividing 1 by that number. But naturally, that's not going to work with the 0.75, so 0.75 will be greater than 20.
So what formula can I apply to any positive number so that they all are less than 1 and greater than 0, and they still sort correctly with this new value (i.e. #1: 20, #2: 1.25, #3: 0.75)
algebra-precalculus elementary-functions
This is a very basic question, I understand that.
So I have these numbers: 20, 0.75, 1.25. If you sort them, it would look like this:
20
1.25
0.75
I need to convert any number that's greater than 1 to a decimal that's less than 1 and greater than 0, and still sort like above.
So I thought about dividing 1 by that number. But naturally, that's not going to work with the 0.75, so 0.75 will be greater than 20.
So what formula can I apply to any positive number so that they all are less than 1 and greater than 0, and they still sort correctly with this new value (i.e. #1: 20, #2: 1.25, #3: 0.75)
algebra-precalculus elementary-functions
edited Aug 7 at 16:35
Mike Pierce
11k93574
11k93574
asked Aug 6 at 14:54
rbhat
1206
1206
Is there a minimum and maximum value that your numbers can take?
â Jared Goguen
Aug 6 at 18:01
3
Part of the question says "any number that's greater than 1," and another part says "apply to any number." You should probably reconcile these two different criteria, especially since the chosen answer fulfills one of them and not the other.
â John
Aug 6 at 18:03
add a comment |Â
Is there a minimum and maximum value that your numbers can take?
â Jared Goguen
Aug 6 at 18:01
3
Part of the question says "any number that's greater than 1," and another part says "apply to any number." You should probably reconcile these two different criteria, especially since the chosen answer fulfills one of them and not the other.
â John
Aug 6 at 18:03
Is there a minimum and maximum value that your numbers can take?
â Jared Goguen
Aug 6 at 18:01
Is there a minimum and maximum value that your numbers can take?
â Jared Goguen
Aug 6 at 18:01
3
3
Part of the question says "any number that's greater than 1," and another part says "apply to any number." You should probably reconcile these two different criteria, especially since the chosen answer fulfills one of them and not the other.
â John
Aug 6 at 18:03
Part of the question says "any number that's greater than 1," and another part says "apply to any number." You should probably reconcile these two different criteria, especially since the chosen answer fulfills one of them and not the other.
â John
Aug 6 at 18:03
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
12
down vote
accepted
There are many approaches. You need an increasing function that has a horizontal asymptote. One simple one is $f(x)=frac x1+x$. When $x=0, f(x)=0$ and as $x$ gets large $f(x)$ goes to $1$.
add a comment |Â
up vote
4
down vote
What about $frac xx+1$? It's an increasing function with values in $[0,1)$.
add a comment |Â
up vote
3
down vote
Try the following function:
$$S(x) = frac 11+e^-x$$
This function has the following graph:
Now, if all of your values are non-negative, you will want to use the following function:
$$S(x) = frac 21+e^-x - 1$$
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
There are many approaches. You need an increasing function that has a horizontal asymptote. One simple one is $f(x)=frac x1+x$. When $x=0, f(x)=0$ and as $x$ gets large $f(x)$ goes to $1$.
add a comment |Â
up vote
12
down vote
accepted
There are many approaches. You need an increasing function that has a horizontal asymptote. One simple one is $f(x)=frac x1+x$. When $x=0, f(x)=0$ and as $x$ gets large $f(x)$ goes to $1$.
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
There are many approaches. You need an increasing function that has a horizontal asymptote. One simple one is $f(x)=frac x1+x$. When $x=0, f(x)=0$ and as $x$ gets large $f(x)$ goes to $1$.
There are many approaches. You need an increasing function that has a horizontal asymptote. One simple one is $f(x)=frac x1+x$. When $x=0, f(x)=0$ and as $x$ gets large $f(x)$ goes to $1$.
answered Aug 6 at 14:58
Ross Millikan
276k21187352
276k21187352
add a comment |Â
add a comment |Â
up vote
4
down vote
What about $frac xx+1$? It's an increasing function with values in $[0,1)$.
add a comment |Â
up vote
4
down vote
What about $frac xx+1$? It's an increasing function with values in $[0,1)$.
add a comment |Â
up vote
4
down vote
up vote
4
down vote
What about $frac xx+1$? It's an increasing function with values in $[0,1)$.
What about $frac xx+1$? It's an increasing function with values in $[0,1)$.
answered Aug 6 at 14:59
Akababa
2,557922
2,557922
add a comment |Â
add a comment |Â
up vote
3
down vote
Try the following function:
$$S(x) = frac 11+e^-x$$
This function has the following graph:
Now, if all of your values are non-negative, you will want to use the following function:
$$S(x) = frac 21+e^-x - 1$$
add a comment |Â
up vote
3
down vote
Try the following function:
$$S(x) = frac 11+e^-x$$
This function has the following graph:
Now, if all of your values are non-negative, you will want to use the following function:
$$S(x) = frac 21+e^-x - 1$$
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Try the following function:
$$S(x) = frac 11+e^-x$$
This function has the following graph:
Now, if all of your values are non-negative, you will want to use the following function:
$$S(x) = frac 21+e^-x - 1$$
Try the following function:
$$S(x) = frac 11+e^-x$$
This function has the following graph:
Now, if all of your values are non-negative, you will want to use the following function:
$$S(x) = frac 21+e^-x - 1$$
answered Aug 6 at 14:59
Rushabh Mehta
1,060114
1,060114
add a comment |Â
add a comment |Â
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Is there a minimum and maximum value that your numbers can take?
â Jared Goguen
Aug 6 at 18:01
3
Part of the question says "any number that's greater than 1," and another part says "apply to any number." You should probably reconcile these two different criteria, especially since the chosen answer fulfills one of them and not the other.
â John
Aug 6 at 18:03