Mixing
Which positive real number x has the property that x, floor of x and x - floor of x form a geometric...
$begingroup$
Which positive real number $x$ has the property that $x$, $lfloor x rfloor$, and $x - lfloor xrfloor$ form a geometric progression (in that order)?
(Recall that $lfloor xrfloor$ means the greatest integer less than or equal to $x$.)
Thanks in advance! Don’t worry about the solution, I just need to put the answer into a program so that I can read the solution
I tried plugging in random numbers but that didn’t work
algebra-precalculus geometric-series
edited Jan 21 at 4:41
gt6989b
34.6k22456
asked Jan 21 at 4:33
MikeyMikey
11
$endgroup$
add a comment |
$begingroup$
Which positive real number $x$ has the property that $x$, $lfloor x rfloor$, and $x - lfloor xrfloor$ form a geometric progression (in that order)?
(Recall that $lfloor xrfloor$ means the greatest integer less than or equal to $x$.)
Thanks in advance! Don’t worry about the solution, I just need to put the answer into a program so that I can read the solution
I tried plugging in random numbers but that didn’t work
algebra-precalculus geometric-series
edited Jan 21 at 4:41
gt6989b
34.6k22456
asked Jan 21 at 4:33
MikeyMikey
11
$endgroup$
1
$begingroup$
Welcome to MSE! Can you please tell us what you have already tried so far? This will help other answerers know what you know and what you need help, and it is generally a helpful way to start off a question. Thank you!
$endgroup$
– BadAtGeometry
Jan 21 at 4:35
$begingroup$
I tried putting in random numbers, but that is about it. I have no clue what to do.
$endgroup$
– Mikey
Jan 21 at 4:38
add a comment |
$begingroup$
Which positive real number $x$ has the property that $x$, $lfloor x rfloor$, and $x - lfloor xrfloor$ form a geometric progression (in that order)?
(Recall that $lfloor xrfloor$ means the greatest integer less than or equal to $x$.)
Thanks in advance! Don’t worry about the solution, I just need to put the answer into a program so that I can read the solution
I tried plugging in random numbers but that didn’t work
algebra-precalculus geometric-series
edited Jan 21 at 4:41
gt6989b
34.6k22456
asked Jan 21 at 4:33
MikeyMikey
11
$endgroup$
Which positive real number $x$ has the property that $x$, $lfloor x rfloor$, and $x - lfloor xrfloor$ form a geometric progression (in that order)?
(Recall that $lfloor xrfloor$ means the greatest integer less than or equal to $x$.)
Thanks in advance! Don’t worry about the solution, I just need to put the answer into a program so that I can read the solution
I tried plugging in random numbers but that didn’t work
algebra-precalculus geometric-series
algebra-precalculus geometric-series
edited Jan 21 at 4:41
gt6989b
34.6k22456
asked Jan 21 at 4:33
MikeyMikey
11
edited Jan 21 at 4:41
gt6989b
34.6k22456
asked Jan 21 at 4:33
MikeyMikey
11
edited Jan 21 at 4:41
gt6989b
34.6k22456
edited Jan 21 at 4:41
gt6989b
34.6k22456
edited Jan 21 at 4:41
gt6989b
34.6k22456
34.6k22456
asked Jan 21 at 4:33
MikeyMikey
11
asked Jan 21 at 4:33
MikeyMikey
11
asked Jan 21 at 4:33
MikeyMikey
11
11
1
$begingroup$
Welcome to MSE! Can you please tell us what you have already tried so far? This will help other answerers know what you know and what you need help, and it is generally a helpful way to start off a question. Thank you!
$endgroup$
– BadAtGeometry
Jan 21 at 4:35
$begingroup$
I tried putting in random numbers, but that is about it. I have no clue what to do.
$endgroup$
– Mikey
Jan 21 at 4:38
add a comment |
1
$begingroup$
Welcome to MSE! Can you please tell us what you have already tried so far? This will help other answerers know what you know and what you need help, and it is generally a helpful way to start off a question. Thank you!
$endgroup$
– BadAtGeometry
Jan 21 at 4:35
$begingroup$
I tried putting in random numbers, but that is about it. I have no clue what to do.
$endgroup$
– Mikey
Jan 21 at 4:38
1
1
$begingroup$
Welcome to MSE! Can you please tell us what you have already tried so far? This will help other answerers know what you know and what you need help, and it is generally a helpful way to start off a question. Thank you!
$endgroup$
– BadAtGeometry
Jan 21 at 4:35
$begingroup$
Welcome to MSE! Can you please tell us what you have already tried so far? This will help other answerers know what you know and what you need help, and it is generally a helpful way to start off a question. Thank you!
$endgroup$
– BadAtGeometry
Jan 21 at 4:35
$begingroup$
I tried putting in random numbers, but that is about it. I have no clue what to do.
$endgroup$
– Mikey
Jan 21 at 4:38
$begingroup$
I tried putting in random numbers, but that is about it. I have no clue what to do.
$endgroup$
– Mikey
Jan 21 at 4:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT
Geometric progression is a sequence where every pair of consecutive terms has the same ratio. Hence, in your case, if $x = n+a$ where $ain[0,1)$ and $n in mathbb{N}$, then your sequence looks like $n+a,n,a$ and that it's geometric implies
$$
frac{n+a}{n} = frac{n}{a}.
$$
Can you finish this?
answered Jan 21 at 4:41
gt6989bgt6989b
34.6k22456
$endgroup$
add a comment |
$begingroup$
Hint:
What the question is saying is that:
$$x=a$$
$$lfloor x rfloor = ar$$
$$x-lfloor x rfloor=ar^2$$
answered Jan 21 at 4:42
Rhys HughesRhys Hughes
6,9571530
$endgroup$
1
$begingroup$
Well in that case, we can add the second and third line to get: $x=ar^2+ar$
$endgroup$
– Mikey
Jan 21 at 4:48
$begingroup$
Better to use $ar^2+ar-a =0 to r =...$
$endgroup$
– Rhys Hughes
Jan 21 at 4:51
1
$begingroup$
We also know that x=a, so r^2+r=1? $r^2+r-1=0$, $r=(-1+-sqrt5)/2. Am I headed the right way?
$endgroup$
– Mikey
Jan 21 at 4:52
1
$begingroup$
@Mikey: Yes. This answer is a good example of reading a word problem and writing equations based on what you are told.
$endgroup$
– Ross Millikan
Jan 21 at 4:58
1
$begingroup$
Thanks! I got it. The answer is $(1+sqrt{5})/2$
$endgroup$
– Mikey
Jan 21 at 5:00
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT
Geometric progression is a sequence where every pair of consecutive terms has the same ratio. Hence, in your case, if $x = n+a$ where $ain[0,1)$ and $n in mathbb{N}$, then your sequence looks like $n+a,n,a$ and that it's geometric implies
$$
frac{n+a}{n} = frac{n}{a}.
$$
Can you finish this?
answered Jan 21 at 4:41
gt6989bgt6989b
34.6k22456
$endgroup$
add a comment |
$begingroup$
HINT
Geometric progression is a sequence where every pair of consecutive terms has the same ratio. Hence, in your case, if $x = n+a$ where $ain[0,1)$ and $n in mathbb{N}$, then your sequence looks like $n+a,n,a$ and that it's geometric implies
$$
frac{n+a}{n} = frac{n}{a}.
$$
Can you finish this?
answered Jan 21 at 4:41
gt6989bgt6989b
34.6k22456
$endgroup$
add a comment |
$begingroup$
HINT
Geometric progression is a sequence where every pair of consecutive terms has the same ratio. Hence, in your case, if $x = n+a$ where $ain[0,1)$ and $n in mathbb{N}$, then your sequence looks like $n+a,n,a$ and that it's geometric implies
$$
frac{n+a}{n} = frac{n}{a}.
$$
Can you finish this?
answered Jan 21 at 4:41
gt6989bgt6989b
34.6k22456
$endgroup$
HINT
Geometric progression is a sequence where every pair of consecutive terms has the same ratio. Hence, in your case, if $x = n+a$ where $ain[0,1)$ and $n in mathbb{N}$, then your sequence looks like $n+a,n,a$ and that it's geometric implies
$$
frac{n+a}{n} = frac{n}{a}.
$$
Can you finish this?
answered Jan 21 at 4:41
gt6989bgt6989b
34.6k22456
answered Jan 21 at 4:41
gt6989bgt6989b
34.6k22456
answered Jan 21 at 4:41
gt6989bgt6989b
34.6k22456
answered Jan 21 at 4:41
gt6989bgt6989b
34.6k22456
34.6k22456
add a comment |
add a comment |
$begingroup$
Hint:
What the question is saying is that:
$$x=a$$
$$lfloor x rfloor = ar$$
$$x-lfloor x rfloor=ar^2$$
answered Jan 21 at 4:42
Rhys HughesRhys Hughes
6,9571530
$endgroup$
1
$begingroup$
Well in that case, we can add the second and third line to get: $x=ar^2+ar$
$endgroup$
– Mikey
Jan 21 at 4:48
$begingroup$
Better to use $ar^2+ar-a =0 to r =...$
$endgroup$
– Rhys Hughes
Jan 21 at 4:51
1
$begingroup$
We also know that x=a, so r^2+r=1? $r^2+r-1=0$, $r=(-1+-sqrt5)/2. Am I headed the right way?
$endgroup$
– Mikey
Jan 21 at 4:52
1
$begingroup$
@Mikey: Yes. This answer is a good example of reading a word problem and writing equations based on what you are told.
$endgroup$
– Ross Millikan
Jan 21 at 4:58
1
$begingroup$
Thanks! I got it. The answer is $(1+sqrt{5})/2$
$endgroup$
– Mikey
Jan 21 at 5:00
|
show 2 more comments
$begingroup$
Hint:
What the question is saying is that:
$$x=a$$
$$lfloor x rfloor = ar$$
$$x-lfloor x rfloor=ar^2$$
answered Jan 21 at 4:42
Rhys HughesRhys Hughes
6,9571530
$endgroup$
1
$begingroup$
Well in that case, we can add the second and third line to get: $x=ar^2+ar$
$endgroup$
– Mikey
Jan 21 at 4:48
$begingroup$
Better to use $ar^2+ar-a =0 to r =...$
$endgroup$
– Rhys Hughes
Jan 21 at 4:51
1
$begingroup$
We also know that x=a, so r^2+r=1? $r^2+r-1=0$, $r=(-1+-sqrt5)/2. Am I headed the right way?
$endgroup$
– Mikey
Jan 21 at 4:52
1
$begingroup$
@Mikey: Yes. This answer is a good example of reading a word problem and writing equations based on what you are told.
$endgroup$
– Ross Millikan
Jan 21 at 4:58
1
$begingroup$
Thanks! I got it. The answer is $(1+sqrt{5})/2$
$endgroup$
– Mikey
Jan 21 at 5:00
|
show 2 more comments
$begingroup$
Hint:
What the question is saying is that:
$$x=a$$
$$lfloor x rfloor = ar$$
$$x-lfloor x rfloor=ar^2$$
answered Jan 21 at 4:42
Rhys HughesRhys Hughes
6,9571530
$endgroup$
Hint:
What the question is saying is that:
$$x=a$$
$$lfloor x rfloor = ar$$
$$x-lfloor x rfloor=ar^2$$
answered Jan 21 at 4:42
Rhys HughesRhys Hughes
6,9571530
answered Jan 21 at 4:42
Rhys HughesRhys Hughes
6,9571530
answered Jan 21 at 4:42
Rhys HughesRhys Hughes
6,9571530
answered Jan 21 at 4:42
Rhys HughesRhys Hughes
6,9571530
6,9571530
1
$begingroup$
Well in that case, we can add the second and third line to get: $x=ar^2+ar$
$endgroup$
– Mikey
Jan 21 at 4:48
$begingroup$
Better to use $ar^2+ar-a =0 to r =...$
$endgroup$
– Rhys Hughes
Jan 21 at 4:51
1
$begingroup$
We also know that x=a, so r^2+r=1? $r^2+r-1=0$, $r=(-1+-sqrt5)/2. Am I headed the right way?
$endgroup$
– Mikey
Jan 21 at 4:52
1
$begingroup$
@Mikey: Yes. This answer is a good example of reading a word problem and writing equations based on what you are told.
$endgroup$
– Ross Millikan
Jan 21 at 4:58
1
$begingroup$
Thanks! I got it. The answer is $(1+sqrt{5})/2$
$endgroup$
– Mikey
Jan 21 at 5:00
|
show 2 more comments
1
$begingroup$
Well in that case, we can add the second and third line to get: $x=ar^2+ar$
$endgroup$
– Mikey
Jan 21 at 4:48
$begingroup$
Better to use $ar^2+ar-a =0 to r =...$
$endgroup$
– Rhys Hughes
Jan 21 at 4:51
1
$begingroup$
We also know that x=a, so r^2+r=1? $r^2+r-1=0$, $r=(-1+-sqrt5)/2. Am I headed the right way?
$endgroup$
– Mikey
Jan 21 at 4:52
1
$begingroup$
@Mikey: Yes. This answer is a good example of reading a word problem and writing equations based on what you are told.
$endgroup$
– Ross Millikan
Jan 21 at 4:58
1
$begingroup$
Thanks! I got it. The answer is $(1+sqrt{5})/2$
$endgroup$
– Mikey
Jan 21 at 5:00
1
1
$begingroup$
Well in that case, we can add the second and third line to get: $x=ar^2+ar$
$endgroup$
– Mikey
Jan 21 at 4:48
$begingroup$
Well in that case, we can add the second and third line to get: $x=ar^2+ar$
$endgroup$
– Mikey
Jan 21 at 4:48
$begingroup$
Better to use $ar^2+ar-a =0 to r =...$
$endgroup$
– Rhys Hughes
Jan 21 at 4:51
$begingroup$
Better to use $ar^2+ar-a =0 to r =...$
$endgroup$
– Rhys Hughes
Jan 21 at 4:51
1
1
$begingroup$
We also know that x=a, so r^2+r=1? $r^2+r-1=0$, $r=(-1+-sqrt5)/2. Am I headed the right way?
$endgroup$
– Mikey
Jan 21 at 4:52
$begingroup$
We also know that x=a, so r^2+r=1? $r^2+r-1=0$, $r=(-1+-sqrt5)/2. Am I headed the right way?
$endgroup$
– Mikey
Jan 21 at 4:52
1
1
$begingroup$
@Mikey: Yes. This answer is a good example of reading a word problem and writing equations based on what you are told.
$endgroup$
– Ross Millikan
Jan 21 at 4:58
$begingroup$
@Mikey: Yes. This answer is a good example of reading a word problem and writing equations based on what you are told.
$endgroup$
– Ross Millikan
Jan 21 at 4:58
1
1
$begingroup$
Thanks! I got it. The answer is $(1+sqrt{5})/2$
$endgroup$
– Mikey
Jan 21 at 5:00
$begingroup$
Thanks! I got it. The answer is $(1+sqrt{5})/2$
$endgroup$
– Mikey
Jan 21 at 5:00
|
show 2 more comments
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$begingroup$
Welcome to MSE! Can you please tell us what you have already tried so far? This will help other answerers know what you know and what you need help, and it is generally a helpful way to start off a question. Thank you!
$endgroup$
– BadAtGeometry
Jan 21 at 4:35
$begingroup$
I tried putting in random numbers, but that is about it. I have no clue what to do.
$endgroup$
– Mikey
Jan 21 at 4:38