Arithmetic laws for rational numbers vs. real numbers
Are there any arithmetic laws that are always true for the set of rational numbers but not always true for the set of real numbers?
This came up because I was doing various exercises in different ways to represent real numbers on a computer (besides the IEEE 754 floating point standard) and accidentally stumbled onto a method that only produces rational numbers. I know that the result of any real number representation will terminate at the length of the data types size and thus the final result of a calculation is always rational, but in most cases the true answer is irrational and then it gets rounded and terminated into being represented as a rational number.
I was curious as to whether this weird property had any benefits that could be taken advantage of.
real-numbers irrational-numbers rational-numbers
asked Nov 26 '18 at 6:53
Rory O'Hare
143
add a comment |
Are there any arithmetic laws that are always true for the set of rational numbers but not always true for the set of real numbers?
This came up because I was doing various exercises in different ways to represent real numbers on a computer (besides the IEEE 754 floating point standard) and accidentally stumbled onto a method that only produces rational numbers. I know that the result of any real number representation will terminate at the length of the data types size and thus the final result of a calculation is always rational, but in most cases the true answer is irrational and then it gets rounded and terminated into being represented as a rational number.
I was curious as to whether this weird property had any benefits that could be taken advantage of.
real-numbers irrational-numbers rational-numbers
asked Nov 26 '18 at 6:53
Rory O'Hare
143
"The final result of a calculation is always rational" - only if you use e.g. fixed-point binary arithmetic. Computer algebra systems like Mathematica are capable of using representations that are much more broad: they can manipulate the number $sqrt{2}$ in complete generality.
– Patrick Stevens
Nov 26 '18 at 7:59
"The equation $x^2=2$ has no solution" is an arithmetic law that is valid in ${mathbb Q}$ but not in ${mathbb R}$. You have to differentiate between numbers as "infinitely precise" mathematical objects (like ${2over7}$ or $pi$) and their (often approximate) representation in a computer.
– Christian Blatter
Nov 26 '18 at 8:09
add a comment |
Are there any arithmetic laws that are always true for the set of rational numbers but not always true for the set of real numbers?
This came up because I was doing various exercises in different ways to represent real numbers on a computer (besides the IEEE 754 floating point standard) and accidentally stumbled onto a method that only produces rational numbers. I know that the result of any real number representation will terminate at the length of the data types size and thus the final result of a calculation is always rational, but in most cases the true answer is irrational and then it gets rounded and terminated into being represented as a rational number.
I was curious as to whether this weird property had any benefits that could be taken advantage of.
real-numbers irrational-numbers rational-numbers
asked Nov 26 '18 at 6:53
Rory O'Hare
143
Are there any arithmetic laws that are always true for the set of rational numbers but not always true for the set of real numbers?
This came up because I was doing various exercises in different ways to represent real numbers on a computer (besides the IEEE 754 floating point standard) and accidentally stumbled onto a method that only produces rational numbers. I know that the result of any real number representation will terminate at the length of the data types size and thus the final result of a calculation is always rational, but in most cases the true answer is irrational and then it gets rounded and terminated into being represented as a rational number.
I was curious as to whether this weird property had any benefits that could be taken advantage of.
real-numbers irrational-numbers rational-numbers
real-numbers irrational-numbers rational-numbers
asked Nov 26 '18 at 6:53
Rory O'Hare
143
asked Nov 26 '18 at 6:53
Rory O'Hare
143
asked Nov 26 '18 at 6:53
Rory O'Hare
143
asked Nov 26 '18 at 6:53
Rory O'Hare
143
asked Nov 26 '18 at 6:53
Rory O'Hare
143
143
"The final result of a calculation is always rational" - only if you use e.g. fixed-point binary arithmetic. Computer algebra systems like Mathematica are capable of using representations that are much more broad: they can manipulate the number $sqrt{2}$ in complete generality.
– Patrick Stevens
Nov 26 '18 at 7:59
"The equation $x^2=2$ has no solution" is an arithmetic law that is valid in ${mathbb Q}$ but not in ${mathbb R}$. You have to differentiate between numbers as "infinitely precise" mathematical objects (like ${2over7}$ or $pi$) and their (often approximate) representation in a computer.
– Christian Blatter
Nov 26 '18 at 8:09
add a comment |
"The final result of a calculation is always rational" - only if you use e.g. fixed-point binary arithmetic. Computer algebra systems like Mathematica are capable of using representations that are much more broad: they can manipulate the number $sqrt{2}$ in complete generality.
– Patrick Stevens
Nov 26 '18 at 7:59
"The equation $x^2=2$ has no solution" is an arithmetic law that is valid in ${mathbb Q}$ but not in ${mathbb R}$. You have to differentiate between numbers as "infinitely precise" mathematical objects (like ${2over7}$ or $pi$) and their (often approximate) representation in a computer.
– Christian Blatter
Nov 26 '18 at 8:09
"The final result of a calculation is always rational" - only if you use e.g. fixed-point binary arithmetic. Computer algebra systems like Mathematica are capable of using representations that are much more broad: they can manipulate the number $sqrt{2}$ in complete generality.
– Patrick Stevens
Nov 26 '18 at 7:59
"The final result of a calculation is always rational" - only if you use e.g. fixed-point binary arithmetic. Computer algebra systems like Mathematica are capable of using representations that are much more broad: they can manipulate the number $sqrt{2}$ in complete generality.
– Patrick Stevens
Nov 26 '18 at 7:59
"The equation $x^2=2$ has no solution" is an arithmetic law that is valid in ${mathbb Q}$ but not in ${mathbb R}$. You have to differentiate between numbers as "infinitely precise" mathematical objects (like ${2over7}$ or $pi$) and their (often approximate) representation in a computer.
– Christian Blatter
Nov 26 '18 at 8:09
"The equation $x^2=2$ has no solution" is an arithmetic law that is valid in ${mathbb Q}$ but not in ${mathbb R}$. You have to differentiate between numbers as "infinitely precise" mathematical objects (like ${2over7}$ or $pi$) and their (often approximate) representation in a computer.
– Christian Blatter
Nov 26 '18 at 8:09
add a comment |
3 Answers
3
active
oldest
votes
Every rational number can be written as $frac ab $ such that $a,b in Bbb{Z} ,bneq 0$ with $operatorname{gcd}(|a|,|b|)=1$.But it's not true for every real number.
answered Nov 26 '18 at 7:28
Thomas Shelby
1,657216
1
Heh, that was the one I knew but I didn't think to mention it :)
– Rory O'Hare
Nov 26 '18 at 7:41
1
Also the product/sum/difference/quotient of two rationals is rational. Of a rational and an irrational is always irrational. But if two irrationals... nothing can be determined.
– fleablood
Nov 26 '18 at 7:43
add a comment |
Given rational non-zero $p, q$ there are non-zero integers $a,b$ where $ap = bq$. That's trivial but imagine two waves with rational period they will have a least common multiple period where they synch up on a periodic bases. A wave with a rational period and one with on odd period will match when the both start but will never synch up again.
(I dunno. That's obvious if you think about it, but it still manages to surprise me. And there's that joke in "Futurama" and the "Channel $sqrt{2}$ News"--- It's funny if you think about it.... At least I think it is funny if you think about it....)
You talk of "using to advantage" and I think the must useful thing would be this sort of period modulo generation that can't happen with irrational values.
answered Nov 26 '18 at 7:55
fleablood
68.3k22685
Do you have "odd period" where you meant "irrational period"?
– badjohn
Nov 26 '18 at 7:58
add a comment |
One of my favourites: floating-point addition is not associative or linear. This very slight failure of linearity turns out to be enough, if stacked as the last layer of a neural network, to allow the network to be trained to do nontrivial work.
answered Nov 26 '18 at 8:02
Patrick Stevens
28.5k52874
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
Every rational number can be written as $frac ab $ such that $a,b in Bbb{Z} ,bneq 0$ with $operatorname{gcd}(|a|,|b|)=1$.But it's not true for every real number.
answered Nov 26 '18 at 7:28
Thomas Shelby
1,657216
1
Heh, that was the one I knew but I didn't think to mention it :)
– Rory O'Hare
Nov 26 '18 at 7:41
1
Also the product/sum/difference/quotient of two rationals is rational. Of a rational and an irrational is always irrational. But if two irrationals... nothing can be determined.
– fleablood
Nov 26 '18 at 7:43
add a comment |
Every rational number can be written as $frac ab $ such that $a,b in Bbb{Z} ,bneq 0$ with $operatorname{gcd}(|a|,|b|)=1$.But it's not true for every real number.
answered Nov 26 '18 at 7:28
Thomas Shelby
1,657216
1
Heh, that was the one I knew but I didn't think to mention it :)
– Rory O'Hare
Nov 26 '18 at 7:41
1
Also the product/sum/difference/quotient of two rationals is rational. Of a rational and an irrational is always irrational. But if two irrationals... nothing can be determined.
– fleablood
Nov 26 '18 at 7:43
add a comment |
Every rational number can be written as $frac ab $ such that $a,b in Bbb{Z} ,bneq 0$ with $operatorname{gcd}(|a|,|b|)=1$.But it's not true for every real number.
answered Nov 26 '18 at 7:28
Thomas Shelby
1,657216
Every rational number can be written as $frac ab $ such that $a,b in Bbb{Z} ,bneq 0$ with $operatorname{gcd}(|a|,|b|)=1$.But it's not true for every real number.
answered Nov 26 '18 at 7:28
Thomas Shelby
1,657216
answered Nov 26 '18 at 7:28
Thomas Shelby
1,657216
answered Nov 26 '18 at 7:28
Thomas Shelby
1,657216
answered Nov 26 '18 at 7:28
Thomas Shelby
1,657216
1,657216
1
Heh, that was the one I knew but I didn't think to mention it :)
– Rory O'Hare
Nov 26 '18 at 7:41
1
Also the product/sum/difference/quotient of two rationals is rational. Of a rational and an irrational is always irrational. But if two irrationals... nothing can be determined.
– fleablood
Nov 26 '18 at 7:43
add a comment |
1
Heh, that was the one I knew but I didn't think to mention it :)
– Rory O'Hare
Nov 26 '18 at 7:41
1
Also the product/sum/difference/quotient of two rationals is rational. Of a rational and an irrational is always irrational. But if two irrationals... nothing can be determined.
– fleablood
Nov 26 '18 at 7:43
1
1
Heh, that was the one I knew but I didn't think to mention it :)
– Rory O'Hare
Nov 26 '18 at 7:41
Heh, that was the one I knew but I didn't think to mention it :)
– Rory O'Hare
Nov 26 '18 at 7:41
1
1
Also the product/sum/difference/quotient of two rationals is rational. Of a rational and an irrational is always irrational. But if two irrationals... nothing can be determined.
– fleablood
Nov 26 '18 at 7:43
Also the product/sum/difference/quotient of two rationals is rational. Of a rational and an irrational is always irrational. But if two irrationals... nothing can be determined.
– fleablood
Nov 26 '18 at 7:43
add a comment |
Given rational non-zero $p, q$ there are non-zero integers $a,b$ where $ap = bq$. That's trivial but imagine two waves with rational period they will have a least common multiple period where they synch up on a periodic bases. A wave with a rational period and one with on odd period will match when the both start but will never synch up again.
(I dunno. That's obvious if you think about it, but it still manages to surprise me. And there's that joke in "Futurama" and the "Channel $sqrt{2}$ News"--- It's funny if you think about it.... At least I think it is funny if you think about it....)
You talk of "using to advantage" and I think the must useful thing would be this sort of period modulo generation that can't happen with irrational values.
answered Nov 26 '18 at 7:55
fleablood
68.3k22685
Do you have "odd period" where you meant "irrational period"?
– badjohn
Nov 26 '18 at 7:58
add a comment |
Given rational non-zero $p, q$ there are non-zero integers $a,b$ where $ap = bq$. That's trivial but imagine two waves with rational period they will have a least common multiple period where they synch up on a periodic bases. A wave with a rational period and one with on odd period will match when the both start but will never synch up again.
(I dunno. That's obvious if you think about it, but it still manages to surprise me. And there's that joke in "Futurama" and the "Channel $sqrt{2}$ News"--- It's funny if you think about it.... At least I think it is funny if you think about it....)
You talk of "using to advantage" and I think the must useful thing would be this sort of period modulo generation that can't happen with irrational values.
answered Nov 26 '18 at 7:55
fleablood
68.3k22685
Do you have "odd period" where you meant "irrational period"?
– badjohn
Nov 26 '18 at 7:58
add a comment |
Given rational non-zero $p, q$ there are non-zero integers $a,b$ where $ap = bq$. That's trivial but imagine two waves with rational period they will have a least common multiple period where they synch up on a periodic bases. A wave with a rational period and one with on odd period will match when the both start but will never synch up again.
(I dunno. That's obvious if you think about it, but it still manages to surprise me. And there's that joke in "Futurama" and the "Channel $sqrt{2}$ News"--- It's funny if you think about it.... At least I think it is funny if you think about it....)
You talk of "using to advantage" and I think the must useful thing would be this sort of period modulo generation that can't happen with irrational values.
answered Nov 26 '18 at 7:55
fleablood
68.3k22685
Given rational non-zero $p, q$ there are non-zero integers $a,b$ where $ap = bq$. That's trivial but imagine two waves with rational period they will have a least common multiple period where they synch up on a periodic bases. A wave with a rational period and one with on odd period will match when the both start but will never synch up again.
(I dunno. That's obvious if you think about it, but it still manages to surprise me. And there's that joke in "Futurama" and the "Channel $sqrt{2}$ News"--- It's funny if you think about it.... At least I think it is funny if you think about it....)
You talk of "using to advantage" and I think the must useful thing would be this sort of period modulo generation that can't happen with irrational values.
answered Nov 26 '18 at 7:55
fleablood
68.3k22685
answered Nov 26 '18 at 7:55
fleablood
68.3k22685
answered Nov 26 '18 at 7:55
fleablood
68.3k22685
answered Nov 26 '18 at 7:55
fleablood
68.3k22685
68.3k22685
Do you have "odd period" where you meant "irrational period"?
– badjohn
Nov 26 '18 at 7:58
add a comment |
Do you have "odd period" where you meant "irrational period"?
– badjohn
Nov 26 '18 at 7:58
Do you have "odd period" where you meant "irrational period"?
– badjohn
Nov 26 '18 at 7:58
Do you have "odd period" where you meant "irrational period"?
– badjohn
Nov 26 '18 at 7:58
add a comment |
One of my favourites: floating-point addition is not associative or linear. This very slight failure of linearity turns out to be enough, if stacked as the last layer of a neural network, to allow the network to be trained to do nontrivial work.
answered Nov 26 '18 at 8:02
Patrick Stevens
28.5k52874
add a comment |
One of my favourites: floating-point addition is not associative or linear. This very slight failure of linearity turns out to be enough, if stacked as the last layer of a neural network, to allow the network to be trained to do nontrivial work.
answered Nov 26 '18 at 8:02
Patrick Stevens
28.5k52874
add a comment |
One of my favourites: floating-point addition is not associative or linear. This very slight failure of linearity turns out to be enough, if stacked as the last layer of a neural network, to allow the network to be trained to do nontrivial work.
answered Nov 26 '18 at 8:02
Patrick Stevens
28.5k52874
One of my favourites: floating-point addition is not associative or linear. This very slight failure of linearity turns out to be enough, if stacked as the last layer of a neural network, to allow the network to be trained to do nontrivial work.
answered Nov 26 '18 at 8:02
Patrick Stevens
28.5k52874
answered Nov 26 '18 at 8:02
Patrick Stevens
28.5k52874
answered Nov 26 '18 at 8:02
Patrick Stevens
28.5k52874
answered Nov 26 '18 at 8:02
Patrick Stevens
28.5k52874
28.5k52874
add a comment |
add a comment |
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"The final result of a calculation is always rational" - only if you use e.g. fixed-point binary arithmetic. Computer algebra systems like Mathematica are capable of using representations that are much more broad: they can manipulate the number $sqrt{2}$ in complete generality.
– Patrick Stevens
Nov 26 '18 at 7:59
"The equation $x^2=2$ has no solution" is an arithmetic law that is valid in ${mathbb Q}$ but not in ${mathbb R}$. You have to differentiate between numbers as "infinitely precise" mathematical objects (like ${2over7}$ or $pi$) and their (often approximate) representation in a computer.
– Christian Blatter
Nov 26 '18 at 8:09