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Between any two powers of $5$ there are either two or three powers of $2$
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Is this statement true?
Between any two consecutive powers of $5$, there are either two or
three powers of $2$.
I can see that this statement is true for cases like
$$5^1 < 2^3 < 2^4 < 5^2$$
or
$$5^3 < 2^7 < 2^8 < 2^9 < 5^4$$
But I am having a trouble figuring out the proof through generalization.
Could somebody help me?
inequality exponentiation
edited Jul 27 at 13:10
Asaf Karagila
291k31402732
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up vote
12
down vote
favorite
Is this statement true?
Between any two consecutive powers of $5$, there are either two or
three powers of $2$.
I can see that this statement is true for cases like
$$5^1 < 2^3 < 2^4 < 5^2$$
or
$$5^3 < 2^7 < 2^8 < 2^9 < 5^4$$
But I am having a trouble figuring out the proof through generalization.
Could somebody help me?
inequality exponentiation
edited Jul 27 at 13:10
Asaf Karagila
291k31402732
19
What's $log_25$?
– Lord Shark the Unknown
Jul 26 at 15:23
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up vote
12
down vote
favorite
up vote
12
down vote
favorite
Is this statement true?
Between any two consecutive powers of $5$, there are either two or
three powers of $2$.
I can see that this statement is true for cases like
$$5^1 < 2^3 < 2^4 < 5^2$$
or
$$5^3 < 2^7 < 2^8 < 2^9 < 5^4$$
But I am having a trouble figuring out the proof through generalization.
Could somebody help me?
inequality exponentiation
edited Jul 27 at 13:10
Asaf Karagila
291k31402732
Is this statement true?
Between any two consecutive powers of $5$, there are either two or
three powers of $2$.
I can see that this statement is true for cases like
$$5^1 < 2^3 < 2^4 < 5^2$$
or
$$5^3 < 2^7 < 2^8 < 2^9 < 5^4$$
But I am having a trouble figuring out the proof through generalization.
Could somebody help me?
inequality exponentiation
edited Jul 27 at 13:10
Asaf Karagila
291k31402732
edited Jul 27 at 13:10
Asaf Karagila
291k31402732
edited Jul 27 at 13:10
Asaf Karagila
291k31402732
edited Jul 27 at 13:10
Asaf Karagila
291k31402732
291k31402732
asked Jul 26 at 15:21
user449415
19
What's $log_25$?
– Lord Shark the Unknown
Jul 26 at 15:23
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19
What's $log_25$?
– Lord Shark the Unknown
Jul 26 at 15:23
19
19
What's $log_25$?
– Lord Shark the Unknown
Jul 26 at 15:23
What's $log_25$?
– Lord Shark the Unknown
Jul 26 at 15:23
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6 Answers
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How many integer multiples of $log_5(2)$ can we fit between the integers $n$ and $n+1$? It's not too hard to see that since $log_5(2)<0.5$, there are at least $2$ such numbers. But, since $3log_5(2)>1$, there are at at most $3$ such numbers. *
So then, if $n$ and $n+1$ straddled $2$ multiples, we would have (for an appropriate integer, $k$): $$n<klog_5(2)<(k+1)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<5^n+1$$ Conversely, if $3$ multiples were straddled, we would have $$n<klog_5(2)<(k+1)log_5(2)<(k+2)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<2^k+2<5^n+1$$
Therefore there are between $2$ and $3$ powers of $2$ between each successive powers of $5$.
* To visualise this, consider the figure below. The red points are multiples of $log_5(2)$. The distance between the red points ($approx0.43$) is small enough to guarantee that at least $2$ of them lie between the black points. However the gap between $4$ red points ($approx1.29$) is too wide to fit between the black points. This is irrespective of where the red points begin.
Here's a fun graph to play around with https://www.desmos.com/calculator/qfqers5mmg. No matter where you start the sequence of red points (by varying the parameter $h$), you inevitably have either $2$ or $3$ red points between the black points.
answered Jul 26 at 16:37
Jam
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Basically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though.
Let $5^a$ and $5^a+1$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^a+1$.
Then we have $2^b-1 leq 5^a < 2^b$ and $2^c < 5^a+1 leq 2^c+1$.
From this we get
$$frac2^c2^b < frac5^a+15^a leq frac2^c+12^b-1 $$
or, equivalently,
$$2^c-b < 5 leq 4 cdot 2^c-b.$$
This inequality can be rewritten as $ frac54 leq 2^c-b < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^a+1$ are either $2^b, 2^b+1 = 2^c$, or $2^b, 2^b+1, 2^b+2 = 2^c.$ This is what you wanted.
answered Jul 26 at 20:06
Dave
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We can see mathematically that there is a ratio between the number of powers of $2$ and the number of powers of $5$ below a certain integer $a$.
Intuitively, the ratio would be $log_25$, or $2.32$. On average, for every power of $5$, you get $2.32$ powers of $2$. That would translate to about $2$ or $3$ powers of $2$ between every two powers of $5$.
If this isn't completely intuitive, notice that there are exactly $3$ more powers of $2$ than $8$, as $log_28 = 3$. For example, out of all integers below $100$, there are $6$ powers of $2$ ($2,4,8,16,32,64$) and $2$ powers of $8$ ($8,64$).
We can extend to this to any integers $x$ and $y$, provided they are not $0$ or $1$: For all integers below an integer $z$, the ratio of the number of exponents of $a$ to the number of exponents of $b$ is $log_ab$.
answered Jul 26 at 15:27
RayDansh
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In the special case $[1,5]$, we see that 1, 2 and 4 are in the interval but 8 and up are not. There are powers of 2 less than and powers of two greater than any other power of 5.
There can't be more than three, because if $0 < a leq b < 2b < 4b < 8b leq 5a$, we get the contradiction $8a leq 5a$ for a positive number $a$.
Nor can there be zero, because if $0 < b < a < 5a < 4b$, we get the contradiction $5b < 4b$ for a positive number b.
Nor can there be one, because the argument still holds if we insert the condition $a leq 2b leq 5a$.
You already found existence proofs for two or three intermediate powers.
The explanations in the other answers are great, but that's a very elementary proof.
answered Jul 26 at 19:48
Davislor
2,082715
Hi, @Davislor. Sorry but I don't quite follow where $8aleq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified?
– Jam
Jul 26 at 20:09
1
@Jam $0< a leq b$, so $8a leq 8b$. Bit, we assumed $8b leq 5a$. Thus the contradiction that $8a leq 5a$ for positive $a$.
– Davislor
Jul 26 at 20:15
@Jam: Do you find my elementary proof any clearer?
– Ilmari Karonen
Jul 27 at 10:33
@IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :)
– Jam
Jul 27 at 11:04
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It's easy enough to see that this is indeed true, even without using logarithms.
Let $a$ be some power of $5$, and let $b$ be the least power of $2$ not less than $a$.
Then the next power of $5$ after $a$ is obviously $5a$, while the next three powers of $2$ after $b$ are $2b$, $4b$ and $8b$.
Since, by definition, $a le b$, it follows that $5a le 5b < 8b$. Thus, there can be at most three powers of $2$ ($b$, $2b$ and $4b$) between $a$ and $5a$.
Conversely, since $b$ is the least power of $2$ not less than $a$, it follows that $frac12 b < a$. Thus, equivalently, $2b < 4a < 5a$, so there must be at least two powers of $2$ ($b$ and $2b$) between $a$ and $5a$.
BTW, if you look closely at the proof above, you may note that it doesn't actually use the assumption that $a$ is a power of $5$ anywhere. Thus, in fact, we've proven a more general result: for any positive number $a$, there are either two or three powers of $2$ between $a$ and $5a$.
(In fact, the proof doesn't really use the assumption that $b$ is a power of $2$, either, so we could generalize the result even further in this direction if we wanted!)
Also, you may notice that the key observation behind the result above is that the number $5$ lies strictly between $2^2 = 4$ and $2^3$ = 8. Thus, by essentially the same logic as above, we can prove a similar result for other bases:
Between any two consecutive powers of $x$ there are at least $k$ and at most $k+1$ powers of $y$, where $x$ and $y$ are any numbers greater than $1$, and $k$ is the largest integer such that $y^k le x$.
answered Jul 27 at 10:31
Ilmari Karonen
19k25180
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We know that
if $y-x >1$, for $xgeq 0, y>0$, then there $exists n in mathbbN, n>0$
s.t. $$x<n<y tag1$$
e.g. $n=left lfloor x right rfloor+1$, because
$$x = left lfloor x right rfloor+x<left lfloor x right rfloor+1<x+1<y$$
In this case $$(k+1)fracln5ln2-kfracln5ln2=fracln5ln2>1$$
and from $(1)$, there $exists ninmathbbN$ s.t.
$$kfracln5ln2<n<(k+1)fracln5ln2 iff\
kln5<nln2<(k+1)ln5 iff \
5^k < 2^n<5^k+1$$
However $fracln5ln2approx 2.32>2$ and $(1)$ can be extended to
if $y-x >2$, for $xgeq 0, y>0$, then there $exists n in mathbbN,n>0$
s.t. $$x<n<n+1<y tag2$$
answered Jul 26 at 16:32
rtybase
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6 Answers
6
active
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6 Answers
6
active
oldest
votes
active
oldest
votes
active
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up vote
23
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How many integer multiples of $log_5(2)$ can we fit between the integers $n$ and $n+1$? It's not too hard to see that since $log_5(2)<0.5$, there are at least $2$ such numbers. But, since $3log_5(2)>1$, there are at at most $3$ such numbers. *
So then, if $n$ and $n+1$ straddled $2$ multiples, we would have (for an appropriate integer, $k$): $$n<klog_5(2)<(k+1)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<5^n+1$$ Conversely, if $3$ multiples were straddled, we would have $$n<klog_5(2)<(k+1)log_5(2)<(k+2)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<2^k+2<5^n+1$$
Therefore there are between $2$ and $3$ powers of $2$ between each successive powers of $5$.
* To visualise this, consider the figure below. The red points are multiples of $log_5(2)$. The distance between the red points ($approx0.43$) is small enough to guarantee that at least $2$ of them lie between the black points. However the gap between $4$ red points ($approx1.29$) is too wide to fit between the black points. This is irrespective of where the red points begin.
Here's a fun graph to play around with https://www.desmos.com/calculator/qfqers5mmg. No matter where you start the sequence of red points (by varying the parameter $h$), you inevitably have either $2$ or $3$ red points between the black points.
answered Jul 26 at 16:37
Jam
4,25211230
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up vote
23
down vote
How many integer multiples of $log_5(2)$ can we fit between the integers $n$ and $n+1$? It's not too hard to see that since $log_5(2)<0.5$, there are at least $2$ such numbers. But, since $3log_5(2)>1$, there are at at most $3$ such numbers. *
So then, if $n$ and $n+1$ straddled $2$ multiples, we would have (for an appropriate integer, $k$): $$n<klog_5(2)<(k+1)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<5^n+1$$ Conversely, if $3$ multiples were straddled, we would have $$n<klog_5(2)<(k+1)log_5(2)<(k+2)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<2^k+2<5^n+1$$
Therefore there are between $2$ and $3$ powers of $2$ between each successive powers of $5$.
* To visualise this, consider the figure below. The red points are multiples of $log_5(2)$. The distance between the red points ($approx0.43$) is small enough to guarantee that at least $2$ of them lie between the black points. However the gap between $4$ red points ($approx1.29$) is too wide to fit between the black points. This is irrespective of where the red points begin.
Here's a fun graph to play around with https://www.desmos.com/calculator/qfqers5mmg. No matter where you start the sequence of red points (by varying the parameter $h$), you inevitably have either $2$ or $3$ red points between the black points.
answered Jul 26 at 16:37
Jam
4,25211230
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up vote
23
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up vote
23
down vote
How many integer multiples of $log_5(2)$ can we fit between the integers $n$ and $n+1$? It's not too hard to see that since $log_5(2)<0.5$, there are at least $2$ such numbers. But, since $3log_5(2)>1$, there are at at most $3$ such numbers. *
So then, if $n$ and $n+1$ straddled $2$ multiples, we would have (for an appropriate integer, $k$): $$n<klog_5(2)<(k+1)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<5^n+1$$ Conversely, if $3$ multiples were straddled, we would have $$n<klog_5(2)<(k+1)log_5(2)<(k+2)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<2^k+2<5^n+1$$
Therefore there are between $2$ and $3$ powers of $2$ between each successive powers of $5$.
* To visualise this, consider the figure below. The red points are multiples of $log_5(2)$. The distance between the red points ($approx0.43$) is small enough to guarantee that at least $2$ of them lie between the black points. However the gap between $4$ red points ($approx1.29$) is too wide to fit between the black points. This is irrespective of where the red points begin.
Here's a fun graph to play around with https://www.desmos.com/calculator/qfqers5mmg. No matter where you start the sequence of red points (by varying the parameter $h$), you inevitably have either $2$ or $3$ red points between the black points.
answered Jul 26 at 16:37
Jam
4,25211230
How many integer multiples of $log_5(2)$ can we fit between the integers $n$ and $n+1$? It's not too hard to see that since $log_5(2)<0.5$, there are at least $2$ such numbers. But, since $3log_5(2)>1$, there are at at most $3$ such numbers. *
So then, if $n$ and $n+1$ straddled $2$ multiples, we would have (for an appropriate integer, $k$): $$n<klog_5(2)<(k+1)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<5^n+1$$ Conversely, if $3$ multiples were straddled, we would have $$n<klog_5(2)<(k+1)log_5(2)<(k+2)log_5(2)<n+1\Rightarrow5^n<2^k<2^k+1<2^k+2<5^n+1$$
Therefore there are between $2$ and $3$ powers of $2$ between each successive powers of $5$.
* To visualise this, consider the figure below. The red points are multiples of $log_5(2)$. The distance between the red points ($approx0.43$) is small enough to guarantee that at least $2$ of them lie between the black points. However the gap between $4$ red points ($approx1.29$) is too wide to fit between the black points. This is irrespective of where the red points begin.
Here's a fun graph to play around with https://www.desmos.com/calculator/qfqers5mmg. No matter where you start the sequence of red points (by varying the parameter $h$), you inevitably have either $2$ or $3$ red points between the black points.
answered Jul 26 at 16:37
Jam
4,25211230
edited Jul 26 at 21:08
answered Jul 26 at 16:37
Jam
4,25211230
answered Jul 26 at 16:37
Jam
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answered Jul 26 at 16:37
Jam
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Basically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though.
Let $5^a$ and $5^a+1$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^a+1$.
Then we have $2^b-1 leq 5^a < 2^b$ and $2^c < 5^a+1 leq 2^c+1$.
From this we get
$$frac2^c2^b < frac5^a+15^a leq frac2^c+12^b-1 $$
or, equivalently,
$$2^c-b < 5 leq 4 cdot 2^c-b.$$
This inequality can be rewritten as $ frac54 leq 2^c-b < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^a+1$ are either $2^b, 2^b+1 = 2^c$, or $2^b, 2^b+1, 2^b+2 = 2^c.$ This is what you wanted.
answered Jul 26 at 20:06
Dave
1262
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up vote
9
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Basically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though.
Let $5^a$ and $5^a+1$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^a+1$.
Then we have $2^b-1 leq 5^a < 2^b$ and $2^c < 5^a+1 leq 2^c+1$.
From this we get
$$frac2^c2^b < frac5^a+15^a leq frac2^c+12^b-1 $$
or, equivalently,
$$2^c-b < 5 leq 4 cdot 2^c-b.$$
This inequality can be rewritten as $ frac54 leq 2^c-b < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^a+1$ are either $2^b, 2^b+1 = 2^c$, or $2^b, 2^b+1, 2^b+2 = 2^c.$ This is what you wanted.
answered Jul 26 at 20:06
Dave
1262
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up vote
9
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up vote
9
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Basically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though.
Let $5^a$ and $5^a+1$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^a+1$.
Then we have $2^b-1 leq 5^a < 2^b$ and $2^c < 5^a+1 leq 2^c+1$.
From this we get
$$frac2^c2^b < frac5^a+15^a leq frac2^c+12^b-1 $$
or, equivalently,
$$2^c-b < 5 leq 4 cdot 2^c-b.$$
This inequality can be rewritten as $ frac54 leq 2^c-b < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^a+1$ are either $2^b, 2^b+1 = 2^c$, or $2^b, 2^b+1, 2^b+2 = 2^c.$ This is what you wanted.
answered Jul 26 at 20:06
Dave
1262
Basically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though.
Let $5^a$ and $5^a+1$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^a+1$.
Then we have $2^b-1 leq 5^a < 2^b$ and $2^c < 5^a+1 leq 2^c+1$.
From this we get
$$frac2^c2^b < frac5^a+15^a leq frac2^c+12^b-1 $$
or, equivalently,
$$2^c-b < 5 leq 4 cdot 2^c-b.$$
This inequality can be rewritten as $ frac54 leq 2^c-b < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^a+1$ are either $2^b, 2^b+1 = 2^c$, or $2^b, 2^b+1, 2^b+2 = 2^c.$ This is what you wanted.
answered Jul 26 at 20:06
Dave
1262
edited Jul 28 at 2:07
answered Jul 26 at 20:06
Dave
1262
answered Jul 26 at 20:06
Dave
1262
answered Jul 26 at 20:06
Dave
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1262
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We can see mathematically that there is a ratio between the number of powers of $2$ and the number of powers of $5$ below a certain integer $a$.
Intuitively, the ratio would be $log_25$, or $2.32$. On average, for every power of $5$, you get $2.32$ powers of $2$. That would translate to about $2$ or $3$ powers of $2$ between every two powers of $5$.
If this isn't completely intuitive, notice that there are exactly $3$ more powers of $2$ than $8$, as $log_28 = 3$. For example, out of all integers below $100$, there are $6$ powers of $2$ ($2,4,8,16,32,64$) and $2$ powers of $8$ ($8,64$).
We can extend to this to any integers $x$ and $y$, provided they are not $0$ or $1$: For all integers below an integer $z$, the ratio of the number of exponents of $a$ to the number of exponents of $b$ is $log_ab$.
answered Jul 26 at 15:27
RayDansh
884214
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up vote
6
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We can see mathematically that there is a ratio between the number of powers of $2$ and the number of powers of $5$ below a certain integer $a$.
Intuitively, the ratio would be $log_25$, or $2.32$. On average, for every power of $5$, you get $2.32$ powers of $2$. That would translate to about $2$ or $3$ powers of $2$ between every two powers of $5$.
If this isn't completely intuitive, notice that there are exactly $3$ more powers of $2$ than $8$, as $log_28 = 3$. For example, out of all integers below $100$, there are $6$ powers of $2$ ($2,4,8,16,32,64$) and $2$ powers of $8$ ($8,64$).
We can extend to this to any integers $x$ and $y$, provided they are not $0$ or $1$: For all integers below an integer $z$, the ratio of the number of exponents of $a$ to the number of exponents of $b$ is $log_ab$.
answered Jul 26 at 15:27
RayDansh
884214
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up vote
6
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up vote
6
down vote
We can see mathematically that there is a ratio between the number of powers of $2$ and the number of powers of $5$ below a certain integer $a$.
Intuitively, the ratio would be $log_25$, or $2.32$. On average, for every power of $5$, you get $2.32$ powers of $2$. That would translate to about $2$ or $3$ powers of $2$ between every two powers of $5$.
If this isn't completely intuitive, notice that there are exactly $3$ more powers of $2$ than $8$, as $log_28 = 3$. For example, out of all integers below $100$, there are $6$ powers of $2$ ($2,4,8,16,32,64$) and $2$ powers of $8$ ($8,64$).
We can extend to this to any integers $x$ and $y$, provided they are not $0$ or $1$: For all integers below an integer $z$, the ratio of the number of exponents of $a$ to the number of exponents of $b$ is $log_ab$.
answered Jul 26 at 15:27
RayDansh
884214
We can see mathematically that there is a ratio between the number of powers of $2$ and the number of powers of $5$ below a certain integer $a$.
Intuitively, the ratio would be $log_25$, or $2.32$. On average, for every power of $5$, you get $2.32$ powers of $2$. That would translate to about $2$ or $3$ powers of $2$ between every two powers of $5$.
If this isn't completely intuitive, notice that there are exactly $3$ more powers of $2$ than $8$, as $log_28 = 3$. For example, out of all integers below $100$, there are $6$ powers of $2$ ($2,4,8,16,32,64$) and $2$ powers of $8$ ($8,64$).
We can extend to this to any integers $x$ and $y$, provided they are not $0$ or $1$: For all integers below an integer $z$, the ratio of the number of exponents of $a$ to the number of exponents of $b$ is $log_ab$.
answered Jul 26 at 15:27
RayDansh
884214
edited Jul 26 at 15:37
answered Jul 26 at 15:27
RayDansh
884214
answered Jul 26 at 15:27
RayDansh
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answered Jul 26 at 15:27
RayDansh
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down vote
In the special case $[1,5]$, we see that 1, 2 and 4 are in the interval but 8 and up are not. There are powers of 2 less than and powers of two greater than any other power of 5.
There can't be more than three, because if $0 < a leq b < 2b < 4b < 8b leq 5a$, we get the contradiction $8a leq 5a$ for a positive number $a$.
Nor can there be zero, because if $0 < b < a < 5a < 4b$, we get the contradiction $5b < 4b$ for a positive number b.
Nor can there be one, because the argument still holds if we insert the condition $a leq 2b leq 5a$.
You already found existence proofs for two or three intermediate powers.
The explanations in the other answers are great, but that's a very elementary proof.
answered Jul 26 at 19:48
Davislor
2,082715
Hi, @Davislor. Sorry but I don't quite follow where $8aleq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified?
– Jam
Jul 26 at 20:09
1
@Jam $0< a leq b$, so $8a leq 8b$. Bit, we assumed $8b leq 5a$. Thus the contradiction that $8a leq 5a$ for positive $a$.
– Davislor
Jul 26 at 20:15
@Jam: Do you find my elementary proof any clearer?
– Ilmari Karonen
Jul 27 at 10:33
@IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :)
– Jam
Jul 27 at 11:04
add a comment |Â
up vote
4
down vote
In the special case $[1,5]$, we see that 1, 2 and 4 are in the interval but 8 and up are not. There are powers of 2 less than and powers of two greater than any other power of 5.
There can't be more than three, because if $0 < a leq b < 2b < 4b < 8b leq 5a$, we get the contradiction $8a leq 5a$ for a positive number $a$.
Nor can there be zero, because if $0 < b < a < 5a < 4b$, we get the contradiction $5b < 4b$ for a positive number b.
Nor can there be one, because the argument still holds if we insert the condition $a leq 2b leq 5a$.
You already found existence proofs for two or three intermediate powers.
The explanations in the other answers are great, but that's a very elementary proof.
answered Jul 26 at 19:48
Davislor
2,082715
Hi, @Davislor. Sorry but I don't quite follow where $8aleq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified?
– Jam
Jul 26 at 20:09
1
@Jam $0< a leq b$, so $8a leq 8b$. Bit, we assumed $8b leq 5a$. Thus the contradiction that $8a leq 5a$ for positive $a$.
– Davislor
Jul 26 at 20:15
@Jam: Do you find my elementary proof any clearer?
– Ilmari Karonen
Jul 27 at 10:33
@IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :)
– Jam
Jul 27 at 11:04
add a comment |Â
up vote
4
down vote
up vote
4
down vote
In the special case $[1,5]$, we see that 1, 2 and 4 are in the interval but 8 and up are not. There are powers of 2 less than and powers of two greater than any other power of 5.
There can't be more than three, because if $0 < a leq b < 2b < 4b < 8b leq 5a$, we get the contradiction $8a leq 5a$ for a positive number $a$.
Nor can there be zero, because if $0 < b < a < 5a < 4b$, we get the contradiction $5b < 4b$ for a positive number b.
Nor can there be one, because the argument still holds if we insert the condition $a leq 2b leq 5a$.
You already found existence proofs for two or three intermediate powers.
The explanations in the other answers are great, but that's a very elementary proof.
answered Jul 26 at 19:48
Davislor
2,082715
In the special case $[1,5]$, we see that 1, 2 and 4 are in the interval but 8 and up are not. There are powers of 2 less than and powers of two greater than any other power of 5.
There can't be more than three, because if $0 < a leq b < 2b < 4b < 8b leq 5a$, we get the contradiction $8a leq 5a$ for a positive number $a$.
Nor can there be zero, because if $0 < b < a < 5a < 4b$, we get the contradiction $5b < 4b$ for a positive number b.
Nor can there be one, because the argument still holds if we insert the condition $a leq 2b leq 5a$.
You already found existence proofs for two or three intermediate powers.
The explanations in the other answers are great, but that's a very elementary proof.
answered Jul 26 at 19:48
Davislor
2,082715
edited Jul 26 at 20:09
answered Jul 26 at 19:48
Davislor
2,082715
answered Jul 26 at 19:48
Davislor
2,082715
answered Jul 26 at 19:48
Davislor
2,082715
2,082715
Hi, @Davislor. Sorry but I don't quite follow where $8aleq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified?
– Jam
Jul 26 at 20:09
1
@Jam $0< a leq b$, so $8a leq 8b$. Bit, we assumed $8b leq 5a$. Thus the contradiction that $8a leq 5a$ for positive $a$.
– Davislor
Jul 26 at 20:15
@Jam: Do you find my elementary proof any clearer?
– Ilmari Karonen
Jul 27 at 10:33
@IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :)
– Jam
Jul 27 at 11:04
add a comment |Â
Hi, @Davislor. Sorry but I don't quite follow where $8aleq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified?
– Jam
Jul 26 at 20:09
1
@Jam $0< a leq b$, so $8a leq 8b$. Bit, we assumed $8b leq 5a$. Thus the contradiction that $8a leq 5a$ for positive $a$.
– Davislor
Jul 26 at 20:15
@Jam: Do you find my elementary proof any clearer?
– Ilmari Karonen
Jul 27 at 10:33
@IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :)
– Jam
Jul 27 at 11:04
Hi, @Davislor. Sorry but I don't quite follow where $8aleq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified?
– Jam
Jul 26 at 20:09
Hi, @Davislor. Sorry but I don't quite follow where $8aleq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified?
– Jam
Jul 26 at 20:09
1
1
@Jam $0< a leq b$, so $8a leq 8b$. Bit, we assumed $8b leq 5a$. Thus the contradiction that $8a leq 5a$ for positive $a$.
– Davislor
Jul 26 at 20:15
@Jam $0< a leq b$, so $8a leq 8b$. Bit, we assumed $8b leq 5a$. Thus the contradiction that $8a leq 5a$ for positive $a$.
– Davislor
Jul 26 at 20:15
@Jam: Do you find my elementary proof any clearer?
– Ilmari Karonen
Jul 27 at 10:33
@Jam: Do you find my elementary proof any clearer?
– Ilmari Karonen
Jul 27 at 10:33
@IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :)
– Jam
Jul 27 at 11:04
@IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :)
– Jam
Jul 27 at 11:04
add a comment |Â
up vote
2
down vote
It's easy enough to see that this is indeed true, even without using logarithms.
Let $a$ be some power of $5$, and let $b$ be the least power of $2$ not less than $a$.
Then the next power of $5$ after $a$ is obviously $5a$, while the next three powers of $2$ after $b$ are $2b$, $4b$ and $8b$.
Since, by definition, $a le b$, it follows that $5a le 5b < 8b$. Thus, there can be at most three powers of $2$ ($b$, $2b$ and $4b$) between $a$ and $5a$.
Conversely, since $b$ is the least power of $2$ not less than $a$, it follows that $frac12 b < a$. Thus, equivalently, $2b < 4a < 5a$, so there must be at least two powers of $2$ ($b$ and $2b$) between $a$ and $5a$.
BTW, if you look closely at the proof above, you may note that it doesn't actually use the assumption that $a$ is a power of $5$ anywhere. Thus, in fact, we've proven a more general result: for any positive number $a$, there are either two or three powers of $2$ between $a$ and $5a$.
(In fact, the proof doesn't really use the assumption that $b$ is a power of $2$, either, so we could generalize the result even further in this direction if we wanted!)
Also, you may notice that the key observation behind the result above is that the number $5$ lies strictly between $2^2 = 4$ and $2^3$ = 8. Thus, by essentially the same logic as above, we can prove a similar result for other bases:
Between any two consecutive powers of $x$ there are at least $k$ and at most $k+1$ powers of $y$, where $x$ and $y$ are any numbers greater than $1$, and $k$ is the largest integer such that $y^k le x$.
answered Jul 27 at 10:31
Ilmari Karonen
19k25180
add a comment |Â
up vote
2
down vote
It's easy enough to see that this is indeed true, even without using logarithms.
Let $a$ be some power of $5$, and let $b$ be the least power of $2$ not less than $a$.
Then the next power of $5$ after $a$ is obviously $5a$, while the next three powers of $2$ after $b$ are $2b$, $4b$ and $8b$.
Since, by definition, $a le b$, it follows that $5a le 5b < 8b$. Thus, there can be at most three powers of $2$ ($b$, $2b$ and $4b$) between $a$ and $5a$.
Conversely, since $b$ is the least power of $2$ not less than $a$, it follows that $frac12 b < a$. Thus, equivalently, $2b < 4a < 5a$, so there must be at least two powers of $2$ ($b$ and $2b$) between $a$ and $5a$.
BTW, if you look closely at the proof above, you may note that it doesn't actually use the assumption that $a$ is a power of $5$ anywhere. Thus, in fact, we've proven a more general result: for any positive number $a$, there are either two or three powers of $2$ between $a$ and $5a$.
(In fact, the proof doesn't really use the assumption that $b$ is a power of $2$, either, so we could generalize the result even further in this direction if we wanted!)
Also, you may notice that the key observation behind the result above is that the number $5$ lies strictly between $2^2 = 4$ and $2^3$ = 8. Thus, by essentially the same logic as above, we can prove a similar result for other bases:
Between any two consecutive powers of $x$ there are at least $k$ and at most $k+1$ powers of $y$, where $x$ and $y$ are any numbers greater than $1$, and $k$ is the largest integer such that $y^k le x$.
answered Jul 27 at 10:31
Ilmari Karonen
19k25180
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's easy enough to see that this is indeed true, even without using logarithms.
Let $a$ be some power of $5$, and let $b$ be the least power of $2$ not less than $a$.
Then the next power of $5$ after $a$ is obviously $5a$, while the next three powers of $2$ after $b$ are $2b$, $4b$ and $8b$.
Since, by definition, $a le b$, it follows that $5a le 5b < 8b$. Thus, there can be at most three powers of $2$ ($b$, $2b$ and $4b$) between $a$ and $5a$.
Conversely, since $b$ is the least power of $2$ not less than $a$, it follows that $frac12 b < a$. Thus, equivalently, $2b < 4a < 5a$, so there must be at least two powers of $2$ ($b$ and $2b$) between $a$ and $5a$.
BTW, if you look closely at the proof above, you may note that it doesn't actually use the assumption that $a$ is a power of $5$ anywhere. Thus, in fact, we've proven a more general result: for any positive number $a$, there are either two or three powers of $2$ between $a$ and $5a$.
(In fact, the proof doesn't really use the assumption that $b$ is a power of $2$, either, so we could generalize the result even further in this direction if we wanted!)
Also, you may notice that the key observation behind the result above is that the number $5$ lies strictly between $2^2 = 4$ and $2^3$ = 8. Thus, by essentially the same logic as above, we can prove a similar result for other bases:
Between any two consecutive powers of $x$ there are at least $k$ and at most $k+1$ powers of $y$, where $x$ and $y$ are any numbers greater than $1$, and $k$ is the largest integer such that $y^k le x$.
answered Jul 27 at 10:31
Ilmari Karonen
19k25180
It's easy enough to see that this is indeed true, even without using logarithms.
Let $a$ be some power of $5$, and let $b$ be the least power of $2$ not less than $a$.
Then the next power of $5$ after $a$ is obviously $5a$, while the next three powers of $2$ after $b$ are $2b$, $4b$ and $8b$.
Since, by definition, $a le b$, it follows that $5a le 5b < 8b$. Thus, there can be at most three powers of $2$ ($b$, $2b$ and $4b$) between $a$ and $5a$.
Conversely, since $b$ is the least power of $2$ not less than $a$, it follows that $frac12 b < a$. Thus, equivalently, $2b < 4a < 5a$, so there must be at least two powers of $2$ ($b$ and $2b$) between $a$ and $5a$.
BTW, if you look closely at the proof above, you may note that it doesn't actually use the assumption that $a$ is a power of $5$ anywhere. Thus, in fact, we've proven a more general result: for any positive number $a$, there are either two or three powers of $2$ between $a$ and $5a$.
(In fact, the proof doesn't really use the assumption that $b$ is a power of $2$, either, so we could generalize the result even further in this direction if we wanted!)
Also, you may notice that the key observation behind the result above is that the number $5$ lies strictly between $2^2 = 4$ and $2^3$ = 8. Thus, by essentially the same logic as above, we can prove a similar result for other bases:
Between any two consecutive powers of $x$ there are at least $k$ and at most $k+1$ powers of $y$, where $x$ and $y$ are any numbers greater than $1$, and $k$ is the largest integer such that $y^k le x$.
answered Jul 27 at 10:31
Ilmari Karonen
19k25180
edited Jul 27 at 10:37
answered Jul 27 at 10:31
Ilmari Karonen
19k25180
answered Jul 27 at 10:31
Ilmari Karonen
19k25180
answered Jul 27 at 10:31
Ilmari Karonen
19k25180
19k25180
add a comment |Â
add a comment |Â
up vote
1
down vote
We know that
if $y-x >1$, for $xgeq 0, y>0$, then there $exists n in mathbbN, n>0$
s.t. $$x<n<y tag1$$
e.g. $n=left lfloor x right rfloor+1$, because
$$x = left lfloor x right rfloor+x<left lfloor x right rfloor+1<x+1<y$$
In this case $$(k+1)fracln5ln2-kfracln5ln2=fracln5ln2>1$$
and from $(1)$, there $exists ninmathbbN$ s.t.
$$kfracln5ln2<n<(k+1)fracln5ln2 iff\
kln5<nln2<(k+1)ln5 iff \
5^k < 2^n<5^k+1$$
However $fracln5ln2approx 2.32>2$ and $(1)$ can be extended to
if $y-x >2$, for $xgeq 0, y>0$, then there $exists n in mathbbN,n>0$
s.t. $$x<n<n+1<y tag2$$
answered Jul 26 at 16:32
rtybase
8,81221333
add a comment |Â
up vote
1
down vote
We know that
if $y-x >1$, for $xgeq 0, y>0$, then there $exists n in mathbbN, n>0$
s.t. $$x<n<y tag1$$
e.g. $n=left lfloor x right rfloor+1$, because
$$x = left lfloor x right rfloor+x<left lfloor x right rfloor+1<x+1<y$$
In this case $$(k+1)fracln5ln2-kfracln5ln2=fracln5ln2>1$$
and from $(1)$, there $exists ninmathbbN$ s.t.
$$kfracln5ln2<n<(k+1)fracln5ln2 iff\
kln5<nln2<(k+1)ln5 iff \
5^k < 2^n<5^k+1$$
However $fracln5ln2approx 2.32>2$ and $(1)$ can be extended to
if $y-x >2$, for $xgeq 0, y>0$, then there $exists n in mathbbN,n>0$
s.t. $$x<n<n+1<y tag2$$
answered Jul 26 at 16:32
rtybase
8,81221333
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We know that
if $y-x >1$, for $xgeq 0, y>0$, then there $exists n in mathbbN, n>0$
s.t. $$x<n<y tag1$$
e.g. $n=left lfloor x right rfloor+1$, because
$$x = left lfloor x right rfloor+x<left lfloor x right rfloor+1<x+1<y$$
In this case $$(k+1)fracln5ln2-kfracln5ln2=fracln5ln2>1$$
and from $(1)$, there $exists ninmathbbN$ s.t.
$$kfracln5ln2<n<(k+1)fracln5ln2 iff\
kln5<nln2<(k+1)ln5 iff \
5^k < 2^n<5^k+1$$
However $fracln5ln2approx 2.32>2$ and $(1)$ can be extended to
if $y-x >2$, for $xgeq 0, y>0$, then there $exists n in mathbbN,n>0$
s.t. $$x<n<n+1<y tag2$$
answered Jul 26 at 16:32
rtybase
8,81221333
We know that
if $y-x >1$, for $xgeq 0, y>0$, then there $exists n in mathbbN, n>0$
s.t. $$x<n<y tag1$$
e.g. $n=left lfloor x right rfloor+1$, because
$$x = left lfloor x right rfloor+x<left lfloor x right rfloor+1<x+1<y$$
In this case $$(k+1)fracln5ln2-kfracln5ln2=fracln5ln2>1$$
and from $(1)$, there $exists ninmathbbN$ s.t.
$$kfracln5ln2<n<(k+1)fracln5ln2 iff\
kln5<nln2<(k+1)ln5 iff \
5^k < 2^n<5^k+1$$
However $fracln5ln2approx 2.32>2$ and $(1)$ can be extended to
if $y-x >2$, for $xgeq 0, y>0$, then there $exists n in mathbbN,n>0$
s.t. $$x<n<n+1<y tag2$$
answered Jul 26 at 16:32
rtybase
8,81221333
edited Jul 26 at 20:57
answered Jul 26 at 16:32
rtybase
8,81221333
answered Jul 26 at 16:32
rtybase
8,81221333
answered Jul 26 at 16:32
rtybase
8,81221333
8,81221333
add a comment |Â
add a comment |Â
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19
What's $log_25$?
– Lord Shark the Unknown
Jul 26 at 15:23