Unique mixed base number representations?
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I would like to represent $pi$ in an arbitary mixed base. Is this possible to do formulaically and uniquely, before knowing the base of the $n^{th}$ digit?
For example, say the number is to be represented by$$A_5B_3C_5D_7E_3dots$$ where the letters are digits and the numbers represent the base of the associated digit. My goal is to formulate a set of rules that puts $mathbb{R}$ into a bijective correspondence with all the possible number representations in the given mixed base. Is this possible?
My first thought is that the $n^{th}$ digit need not represent some $n^{th}$ power of $frac 1{b}$ else there will be unrepresentable numbers.
Edit: the purpose of expressing a number this way is to enable a digit walk on a non-uniform tessellation graph. For example, expressing $pi$ in base four allows for a digit walk on a square lattice. Well, what if you want to do a digit walk on the edge graph of an arbitrary tiling of the plane (without uniform degree)?
real-analysis real-numbers analytic-number-theory
asked Nov 21 at 5:29
David Diaz
851419
add a comment |
up vote
1
down vote
favorite
I would like to represent $pi$ in an arbitary mixed base. Is this possible to do formulaically and uniquely, before knowing the base of the $n^{th}$ digit?
For example, say the number is to be represented by$$A_5B_3C_5D_7E_3dots$$ where the letters are digits and the numbers represent the base of the associated digit. My goal is to formulate a set of rules that puts $mathbb{R}$ into a bijective correspondence with all the possible number representations in the given mixed base. Is this possible?
My first thought is that the $n^{th}$ digit need not represent some $n^{th}$ power of $frac 1{b}$ else there will be unrepresentable numbers.
Edit: the purpose of expressing a number this way is to enable a digit walk on a non-uniform tessellation graph. For example, expressing $pi$ in base four allows for a digit walk on a square lattice. Well, what if you want to do a digit walk on the edge graph of an arbitrary tiling of the plane (without uniform degree)?
real-analysis real-numbers analytic-number-theory
asked Nov 21 at 5:29
David Diaz
851419
1
Pick a sequence of integers $b_j ge 2$. Then any real number is of the form $a_0 + sum_{j=1}^infty frac{a_j}{prod_{l=1}^j b_l}$ with $a_0 in mathbb{Z}, a_j in 0 ldots b_j-1$. That representation is non-unique as $sum_{j=J}^infty frac{b_j-1}{prod_{l=1}^j b_l} = frac{1}{prod_{l=1}^{J-1} b_l}$
– reuns
Nov 21 at 17:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I would like to represent $pi$ in an arbitary mixed base. Is this possible to do formulaically and uniquely, before knowing the base of the $n^{th}$ digit?
For example, say the number is to be represented by$$A_5B_3C_5D_7E_3dots$$ where the letters are digits and the numbers represent the base of the associated digit. My goal is to formulate a set of rules that puts $mathbb{R}$ into a bijective correspondence with all the possible number representations in the given mixed base. Is this possible?
My first thought is that the $n^{th}$ digit need not represent some $n^{th}$ power of $frac 1{b}$ else there will be unrepresentable numbers.
Edit: the purpose of expressing a number this way is to enable a digit walk on a non-uniform tessellation graph. For example, expressing $pi$ in base four allows for a digit walk on a square lattice. Well, what if you want to do a digit walk on the edge graph of an arbitrary tiling of the plane (without uniform degree)?
real-analysis real-numbers analytic-number-theory
asked Nov 21 at 5:29
David Diaz
851419
I would like to represent $pi$ in an arbitary mixed base. Is this possible to do formulaically and uniquely, before knowing the base of the $n^{th}$ digit?
For example, say the number is to be represented by$$A_5B_3C_5D_7E_3dots$$ where the letters are digits and the numbers represent the base of the associated digit. My goal is to formulate a set of rules that puts $mathbb{R}$ into a bijective correspondence with all the possible number representations in the given mixed base. Is this possible?
My first thought is that the $n^{th}$ digit need not represent some $n^{th}$ power of $frac 1{b}$ else there will be unrepresentable numbers.
Edit: the purpose of expressing a number this way is to enable a digit walk on a non-uniform tessellation graph. For example, expressing $pi$ in base four allows for a digit walk on a square lattice. Well, what if you want to do a digit walk on the edge graph of an arbitrary tiling of the plane (without uniform degree)?
real-analysis real-numbers analytic-number-theory
real-analysis real-numbers analytic-number-theory
asked Nov 21 at 5:29
David Diaz
851419
asked Nov 21 at 5:29
David Diaz
851419
edited Nov 23 at 10:31
asked Nov 21 at 5:29
David Diaz
851419
asked Nov 21 at 5:29
David Diaz
851419
asked Nov 21 at 5:29
David Diaz
851419
851419
1
Pick a sequence of integers $b_j ge 2$. Then any real number is of the form $a_0 + sum_{j=1}^infty frac{a_j}{prod_{l=1}^j b_l}$ with $a_0 in mathbb{Z}, a_j in 0 ldots b_j-1$. That representation is non-unique as $sum_{j=J}^infty frac{b_j-1}{prod_{l=1}^j b_l} = frac{1}{prod_{l=1}^{J-1} b_l}$
– reuns
Nov 21 at 17:36
add a comment |
1
Pick a sequence of integers $b_j ge 2$. Then any real number is of the form $a_0 + sum_{j=1}^infty frac{a_j}{prod_{l=1}^j b_l}$ with $a_0 in mathbb{Z}, a_j in 0 ldots b_j-1$. That representation is non-unique as $sum_{j=J}^infty frac{b_j-1}{prod_{l=1}^j b_l} = frac{1}{prod_{l=1}^{J-1} b_l}$
– reuns
Nov 21 at 17:36
1
1
Pick a sequence of integers $b_j ge 2$. Then any real number is of the form $a_0 + sum_{j=1}^infty frac{a_j}{prod_{l=1}^j b_l}$ with $a_0 in mathbb{Z}, a_j in 0 ldots b_j-1$. That representation is non-unique as $sum_{j=J}^infty frac{b_j-1}{prod_{l=1}^j b_l} = frac{1}{prod_{l=1}^{J-1} b_l}$
– reuns
Nov 21 at 17:36
Pick a sequence of integers $b_j ge 2$. Then any real number is of the form $a_0 + sum_{j=1}^infty frac{a_j}{prod_{l=1}^j b_l}$ with $a_0 in mathbb{Z}, a_j in 0 ldots b_j-1$. That representation is non-unique as $sum_{j=J}^infty frac{b_j-1}{prod_{l=1}^j b_l} = frac{1}{prod_{l=1}^{J-1} b_l}$
– reuns
Nov 21 at 17:36
add a comment |
1 Answer
1
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up vote
2
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Yes you can. Let us consider representing $pi$ as $3.A_5B_3C_5ldots$ We can use the leading $3$ as long as the base in the ones place is at least $4$. Any digit in the place after the radix point is worth $frac 15$ and $pi lt 3frac 15$, so $A=0$. Another way to see that is we multiply the fractional part by $5$ and take the integer part of the result. As $5 cdot 0.14159265 =0.7796325$, which is less than $1$ we again see $A=0$. Now since the next base is $3$ we multiply by $3$ and take the integer part. $3 cdot 0.7796325=2.3388675$ so $B=2$ and we strip that off to continue. Now $5 cdot 0.3388675=1.6943375$, so $C=1$ and we have $pi approx 3.0_52_31_5ldots$ You have an unambiguous representation as long as you use digits less than the base in each place unless the expression would terminate. The $1$ in the third place represents $frac 1{5cdot 3 cdot 5}=frac 1{75}$, so we have $pi approx 3+frac 2{15}+frac 1{75}+ldots$. You can do this, but I don't see why you would want to. The point of numeric representations is to understand numbers or to compute with them. This makes both hard.
answered Nov 21 at 6:15
Ross Millikan
288k23195366
Thanks Ross, exactly what I need. Small typo: B was initially 2, then 1 in the rest of the post.
– David Diaz
Nov 21 at 17:16
@DavidDiaz: I have fixed it. Thanks
– Ross Millikan
Nov 21 at 17:58
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes you can. Let us consider representing $pi$ as $3.A_5B_3C_5ldots$ We can use the leading $3$ as long as the base in the ones place is at least $4$. Any digit in the place after the radix point is worth $frac 15$ and $pi lt 3frac 15$, so $A=0$. Another way to see that is we multiply the fractional part by $5$ and take the integer part of the result. As $5 cdot 0.14159265 =0.7796325$, which is less than $1$ we again see $A=0$. Now since the next base is $3$ we multiply by $3$ and take the integer part. $3 cdot 0.7796325=2.3388675$ so $B=2$ and we strip that off to continue. Now $5 cdot 0.3388675=1.6943375$, so $C=1$ and we have $pi approx 3.0_52_31_5ldots$ You have an unambiguous representation as long as you use digits less than the base in each place unless the expression would terminate. The $1$ in the third place represents $frac 1{5cdot 3 cdot 5}=frac 1{75}$, so we have $pi approx 3+frac 2{15}+frac 1{75}+ldots$. You can do this, but I don't see why you would want to. The point of numeric representations is to understand numbers or to compute with them. This makes both hard.
answered Nov 21 at 6:15
Ross Millikan
288k23195366
Thanks Ross, exactly what I need. Small typo: B was initially 2, then 1 in the rest of the post.
– David Diaz
Nov 21 at 17:16
@DavidDiaz: I have fixed it. Thanks
– Ross Millikan
Nov 21 at 17:58
add a comment |
up vote
2
down vote
accepted
Yes you can. Let us consider representing $pi$ as $3.A_5B_3C_5ldots$ We can use the leading $3$ as long as the base in the ones place is at least $4$. Any digit in the place after the radix point is worth $frac 15$ and $pi lt 3frac 15$, so $A=0$. Another way to see that is we multiply the fractional part by $5$ and take the integer part of the result. As $5 cdot 0.14159265 =0.7796325$, which is less than $1$ we again see $A=0$. Now since the next base is $3$ we multiply by $3$ and take the integer part. $3 cdot 0.7796325=2.3388675$ so $B=2$ and we strip that off to continue. Now $5 cdot 0.3388675=1.6943375$, so $C=1$ and we have $pi approx 3.0_52_31_5ldots$ You have an unambiguous representation as long as you use digits less than the base in each place unless the expression would terminate. The $1$ in the third place represents $frac 1{5cdot 3 cdot 5}=frac 1{75}$, so we have $pi approx 3+frac 2{15}+frac 1{75}+ldots$. You can do this, but I don't see why you would want to. The point of numeric representations is to understand numbers or to compute with them. This makes both hard.
answered Nov 21 at 6:15
Ross Millikan
288k23195366
Thanks Ross, exactly what I need. Small typo: B was initially 2, then 1 in the rest of the post.
– David Diaz
Nov 21 at 17:16
@DavidDiaz: I have fixed it. Thanks
– Ross Millikan
Nov 21 at 17:58
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes you can. Let us consider representing $pi$ as $3.A_5B_3C_5ldots$ We can use the leading $3$ as long as the base in the ones place is at least $4$. Any digit in the place after the radix point is worth $frac 15$ and $pi lt 3frac 15$, so $A=0$. Another way to see that is we multiply the fractional part by $5$ and take the integer part of the result. As $5 cdot 0.14159265 =0.7796325$, which is less than $1$ we again see $A=0$. Now since the next base is $3$ we multiply by $3$ and take the integer part. $3 cdot 0.7796325=2.3388675$ so $B=2$ and we strip that off to continue. Now $5 cdot 0.3388675=1.6943375$, so $C=1$ and we have $pi approx 3.0_52_31_5ldots$ You have an unambiguous representation as long as you use digits less than the base in each place unless the expression would terminate. The $1$ in the third place represents $frac 1{5cdot 3 cdot 5}=frac 1{75}$, so we have $pi approx 3+frac 2{15}+frac 1{75}+ldots$. You can do this, but I don't see why you would want to. The point of numeric representations is to understand numbers or to compute with them. This makes both hard.
answered Nov 21 at 6:15
Ross Millikan
288k23195366
Yes you can. Let us consider representing $pi$ as $3.A_5B_3C_5ldots$ We can use the leading $3$ as long as the base in the ones place is at least $4$. Any digit in the place after the radix point is worth $frac 15$ and $pi lt 3frac 15$, so $A=0$. Another way to see that is we multiply the fractional part by $5$ and take the integer part of the result. As $5 cdot 0.14159265 =0.7796325$, which is less than $1$ we again see $A=0$. Now since the next base is $3$ we multiply by $3$ and take the integer part. $3 cdot 0.7796325=2.3388675$ so $B=2$ and we strip that off to continue. Now $5 cdot 0.3388675=1.6943375$, so $C=1$ and we have $pi approx 3.0_52_31_5ldots$ You have an unambiguous representation as long as you use digits less than the base in each place unless the expression would terminate. The $1$ in the third place represents $frac 1{5cdot 3 cdot 5}=frac 1{75}$, so we have $pi approx 3+frac 2{15}+frac 1{75}+ldots$. You can do this, but I don't see why you would want to. The point of numeric representations is to understand numbers or to compute with them. This makes both hard.
answered Nov 21 at 6:15
Ross Millikan
288k23195366
edited Nov 21 at 17:56
answered Nov 21 at 6:15
Ross Millikan
288k23195366
answered Nov 21 at 6:15
Ross Millikan
288k23195366
answered Nov 21 at 6:15
Ross Millikan
288k23195366
288k23195366
Thanks Ross, exactly what I need. Small typo: B was initially 2, then 1 in the rest of the post.
– David Diaz
Nov 21 at 17:16
@DavidDiaz: I have fixed it. Thanks
– Ross Millikan
Nov 21 at 17:58
add a comment |
Thanks Ross, exactly what I need. Small typo: B was initially 2, then 1 in the rest of the post.
– David Diaz
Nov 21 at 17:16
@DavidDiaz: I have fixed it. Thanks
– Ross Millikan
Nov 21 at 17:58
Thanks Ross, exactly what I need. Small typo: B was initially 2, then 1 in the rest of the post.
– David Diaz
Nov 21 at 17:16
Thanks Ross, exactly what I need. Small typo: B was initially 2, then 1 in the rest of the post.
– David Diaz
Nov 21 at 17:16
@DavidDiaz: I have fixed it. Thanks
– Ross Millikan
Nov 21 at 17:58
@DavidDiaz: I have fixed it. Thanks
– Ross Millikan
Nov 21 at 17:58
add a comment |
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Pick a sequence of integers $b_j ge 2$. Then any real number is of the form $a_0 + sum_{j=1}^infty frac{a_j}{prod_{l=1}^j b_l}$ with $a_0 in mathbb{Z}, a_j in 0 ldots b_j-1$. That representation is non-unique as $sum_{j=J}^infty frac{b_j-1}{prod_{l=1}^j b_l} = frac{1}{prod_{l=1}^{J-1} b_l}$
– reuns
Nov 21 at 17:36