Order of operations involving mod (%) The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Order of operations in evaluating a polynomialconfused about order of operationsThe Order of Operations with BracketsHow do I solve operations involving fractional surds?Transformations of Functions - What are the Order of Operations?Order of Inverse OperationsOrder of math operationsOrder of operations when using the ÷ symbol.Simplifying Order of OperationsBasic error in order of operations
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Order of operations involving mod (%)
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Order of operations in evaluating a polynomialconfused about order of operationsThe Order of Operations with BracketsHow do I solve operations involving fractional surds?Transformations of Functions - What are the Order of Operations?Order of Inverse OperationsOrder of math operationsOrder of operations when using the ÷ symbol.Simplifying Order of OperationsBasic error in order of operations
$begingroup$
I was taking a programming test last night that had a math equation that simplified to 11 % 2 * 3, no () or likewise. When I compute it, being taught modulous occurs at the same level of multiplication or division. As a result I get
11 % 2 * 3
1 * 3
3
Final result I get is 3. When I checked my math in spotlight calculator on my mac I get 5. And then we jump into the rabbit hole.
I go to my preferred math calculator and visit WolframAlpha. Answer is 5.
Google the formula and get 3.
So we have two possible ways to handle this equation,
(11%2)*3 = 3
11%(2*3) = 5
Which way is correct and definitive? I need sources as if Wolfram alpha is doing incorrectly, I would like to have them change it.
algebra-precalculus
asked Apr 24 '14 at 16:54
traisjamestraisjames
1313
$endgroup$
add a comment |
$begingroup$
I was taking a programming test last night that had a math equation that simplified to 11 % 2 * 3, no () or likewise. When I compute it, being taught modulous occurs at the same level of multiplication or division. As a result I get
11 % 2 * 3
1 * 3
3
Final result I get is 3. When I checked my math in spotlight calculator on my mac I get 5. And then we jump into the rabbit hole.
I go to my preferred math calculator and visit WolframAlpha. Answer is 5.
Google the formula and get 3.
So we have two possible ways to handle this equation,
(11%2)*3 = 3
11%(2*3) = 5
Which way is correct and definitive? I need sources as if Wolfram alpha is doing incorrectly, I would like to have them change it.
algebra-precalculus
asked Apr 24 '14 at 16:54
traisjamestraisjames
1313
$endgroup$
2
$begingroup$
There might be differences in the correct order of operations for mathematics and for various programming languages.
$endgroup$
– naslundx
Apr 24 '14 at 16:57
$begingroup$
For cases like this, where there is no ubiquitous convention, the denotation is ambiguous, so you should always use parentheses to disambiguate.
$endgroup$
– Bill Dubuque
Apr 24 '14 at 17:08
$begingroup$
I would but with a test question they don't alway do stuff like that.
$endgroup$
– traisjames
Apr 24 '14 at 17:42
1
$begingroup$
@naslundx: examples show the order for Google is not the same as for the other softwares tested. Which is order "for mathematics"? $amod bcdot c=amod (bcdot c)$ or $amod bcdot c=(amod b)cdot c$? Shouldn't this be the answer?
$endgroup$
– MattAllegro
Apr 24 '14 at 19:25
add a comment |
$begingroup$
I was taking a programming test last night that had a math equation that simplified to 11 % 2 * 3, no () or likewise. When I compute it, being taught modulous occurs at the same level of multiplication or division. As a result I get
11 % 2 * 3
1 * 3
3
Final result I get is 3. When I checked my math in spotlight calculator on my mac I get 5. And then we jump into the rabbit hole.
I go to my preferred math calculator and visit WolframAlpha. Answer is 5.
Google the formula and get 3.
So we have two possible ways to handle this equation,
(11%2)*3 = 3
11%(2*3) = 5
Which way is correct and definitive? I need sources as if Wolfram alpha is doing incorrectly, I would like to have them change it.
algebra-precalculus
asked Apr 24 '14 at 16:54
traisjamestraisjames
1313
$endgroup$
I was taking a programming test last night that had a math equation that simplified to 11 % 2 * 3, no () or likewise. When I compute it, being taught modulous occurs at the same level of multiplication or division. As a result I get
11 % 2 * 3
1 * 3
3
Final result I get is 3. When I checked my math in spotlight calculator on my mac I get 5. And then we jump into the rabbit hole.
I go to my preferred math calculator and visit WolframAlpha. Answer is 5.
Google the formula and get 3.
So we have two possible ways to handle this equation,
(11%2)*3 = 3
11%(2*3) = 5
Which way is correct and definitive? I need sources as if Wolfram alpha is doing incorrectly, I would like to have them change it.
algebra-precalculus
algebra-precalculus
asked Apr 24 '14 at 16:54
traisjamestraisjames
1313
asked Apr 24 '14 at 16:54
traisjamestraisjames
1313
asked Apr 24 '14 at 16:54
traisjamestraisjames
1313
asked Apr 24 '14 at 16:54
traisjamestraisjames
1313
asked Apr 24 '14 at 16:54
traisjamestraisjames
1313
1313
2
$begingroup$
There might be differences in the correct order of operations for mathematics and for various programming languages.
$endgroup$
– naslundx
Apr 24 '14 at 16:57
$begingroup$
For cases like this, where there is no ubiquitous convention, the denotation is ambiguous, so you should always use parentheses to disambiguate.
$endgroup$
– Bill Dubuque
Apr 24 '14 at 17:08
$begingroup$
I would but with a test question they don't alway do stuff like that.
$endgroup$
– traisjames
Apr 24 '14 at 17:42
1
$begingroup$
@naslundx: examples show the order for Google is not the same as for the other softwares tested. Which is order "for mathematics"? $amod bcdot c=amod (bcdot c)$ or $amod bcdot c=(amod b)cdot c$? Shouldn't this be the answer?
$endgroup$
– MattAllegro
Apr 24 '14 at 19:25
add a comment |
2
$begingroup$
There might be differences in the correct order of operations for mathematics and for various programming languages.
$endgroup$
– naslundx
Apr 24 '14 at 16:57
$begingroup$
For cases like this, where there is no ubiquitous convention, the denotation is ambiguous, so you should always use parentheses to disambiguate.
$endgroup$
– Bill Dubuque
Apr 24 '14 at 17:08
$begingroup$
I would but with a test question they don't alway do stuff like that.
$endgroup$
– traisjames
Apr 24 '14 at 17:42
1
$begingroup$
@naslundx: examples show the order for Google is not the same as for the other softwares tested. Which is order "for mathematics"? $amod bcdot c=amod (bcdot c)$ or $amod bcdot c=(amod b)cdot c$? Shouldn't this be the answer?
$endgroup$
– MattAllegro
Apr 24 '14 at 19:25
2
2
$begingroup$
There might be differences in the correct order of operations for mathematics and for various programming languages.
$endgroup$
– naslundx
Apr 24 '14 at 16:57
$begingroup$
There might be differences in the correct order of operations for mathematics and for various programming languages.
$endgroup$
– naslundx
Apr 24 '14 at 16:57
$begingroup$
For cases like this, where there is no ubiquitous convention, the denotation is ambiguous, so you should always use parentheses to disambiguate.
$endgroup$
– Bill Dubuque
Apr 24 '14 at 17:08
$begingroup$
For cases like this, where there is no ubiquitous convention, the denotation is ambiguous, so you should always use parentheses to disambiguate.
$endgroup$
– Bill Dubuque
Apr 24 '14 at 17:08
$begingroup$
I would but with a test question they don't alway do stuff like that.
$endgroup$
– traisjames
Apr 24 '14 at 17:42
$begingroup$
I would but with a test question they don't alway do stuff like that.
$endgroup$
– traisjames
Apr 24 '14 at 17:42
1
1
$begingroup$
@naslundx: examples show the order for Google is not the same as for the other softwares tested. Which is order "for mathematics"? $amod bcdot c=amod (bcdot c)$ or $amod bcdot c=(amod b)cdot c$? Shouldn't this be the answer?
$endgroup$
– MattAllegro
Apr 24 '14 at 19:25
$begingroup$
@naslundx: examples show the order for Google is not the same as for the other softwares tested. Which is order "for mathematics"? $amod bcdot c=amod (bcdot c)$ or $amod bcdot c=(amod b)cdot c$? Shouldn't this be the answer?
$endgroup$
– MattAllegro
Apr 24 '14 at 19:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is no well established convention for the order of precedence between "the modulus operator" and multiplication (or addition, for that matter, because the same problem arises if you replace multiplication with addition in your question.) You have in your hands an example demonstrating that they can't be used with equal precedence.
The acceptability of the final answer hinges entirely upon the choice made for precedence.
While % can certainly be treated as a binary operation of natural numbers, it is not really the main way mathematicians use modulus. Rather than looking at it as an operation, we think of it as indicating the 'environment' where arithmetic is taking place. So $pmod n$ indicates that we are not working in the natural numbers but with a quotient ring of the integers.
In that context, you can establish an identity between modulus and $+/*$ operations. You could state it this way:
$$(a+b)%n = ((a %n)+(b%n)) %n$$
and
$$(a*b)%n = ((a %n)*(b%n)) %n$$
answered Oct 1 '15 at 10:37
rschwiebrschwieb
108k12104253
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
There is no well established convention for the order of precedence between "the modulus operator" and multiplication (or addition, for that matter, because the same problem arises if you replace multiplication with addition in your question.) You have in your hands an example demonstrating that they can't be used with equal precedence.
The acceptability of the final answer hinges entirely upon the choice made for precedence.
While % can certainly be treated as a binary operation of natural numbers, it is not really the main way mathematicians use modulus. Rather than looking at it as an operation, we think of it as indicating the 'environment' where arithmetic is taking place. So $pmod n$ indicates that we are not working in the natural numbers but with a quotient ring of the integers.
In that context, you can establish an identity between modulus and $+/*$ operations. You could state it this way:
$$(a+b)%n = ((a %n)+(b%n)) %n$$
and
$$(a*b)%n = ((a %n)*(b%n)) %n$$
answered Oct 1 '15 at 10:37
rschwiebrschwieb
108k12104253
$endgroup$
add a comment |
$begingroup$
There is no well established convention for the order of precedence between "the modulus operator" and multiplication (or addition, for that matter, because the same problem arises if you replace multiplication with addition in your question.) You have in your hands an example demonstrating that they can't be used with equal precedence.
The acceptability of the final answer hinges entirely upon the choice made for precedence.
While % can certainly be treated as a binary operation of natural numbers, it is not really the main way mathematicians use modulus. Rather than looking at it as an operation, we think of it as indicating the 'environment' where arithmetic is taking place. So $pmod n$ indicates that we are not working in the natural numbers but with a quotient ring of the integers.
In that context, you can establish an identity between modulus and $+/*$ operations. You could state it this way:
$$(a+b)%n = ((a %n)+(b%n)) %n$$
and
$$(a*b)%n = ((a %n)*(b%n)) %n$$
answered Oct 1 '15 at 10:37
rschwiebrschwieb
108k12104253
$endgroup$
add a comment |
$begingroup$
There is no well established convention for the order of precedence between "the modulus operator" and multiplication (or addition, for that matter, because the same problem arises if you replace multiplication with addition in your question.) You have in your hands an example demonstrating that they can't be used with equal precedence.
The acceptability of the final answer hinges entirely upon the choice made for precedence.
While % can certainly be treated as a binary operation of natural numbers, it is not really the main way mathematicians use modulus. Rather than looking at it as an operation, we think of it as indicating the 'environment' where arithmetic is taking place. So $pmod n$ indicates that we are not working in the natural numbers but with a quotient ring of the integers.
In that context, you can establish an identity between modulus and $+/*$ operations. You could state it this way:
$$(a+b)%n = ((a %n)+(b%n)) %n$$
and
$$(a*b)%n = ((a %n)*(b%n)) %n$$
answered Oct 1 '15 at 10:37
rschwiebrschwieb
108k12104253
$endgroup$
There is no well established convention for the order of precedence between "the modulus operator" and multiplication (or addition, for that matter, because the same problem arises if you replace multiplication with addition in your question.) You have in your hands an example demonstrating that they can't be used with equal precedence.
The acceptability of the final answer hinges entirely upon the choice made for precedence.
While % can certainly be treated as a binary operation of natural numbers, it is not really the main way mathematicians use modulus. Rather than looking at it as an operation, we think of it as indicating the 'environment' where arithmetic is taking place. So $pmod n$ indicates that we are not working in the natural numbers but with a quotient ring of the integers.
In that context, you can establish an identity between modulus and $+/*$ operations. You could state it this way:
$$(a+b)%n = ((a %n)+(b%n)) %n$$
and
$$(a*b)%n = ((a %n)*(b%n)) %n$$
answered Oct 1 '15 at 10:37
rschwiebrschwieb
108k12104253
answered Oct 1 '15 at 10:37
rschwiebrschwieb
108k12104253
answered Oct 1 '15 at 10:37
rschwiebrschwieb
108k12104253
answered Oct 1 '15 at 10:37
rschwiebrschwieb
108k12104253
108k12104253
add a comment |
add a comment |
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2
$begingroup$
There might be differences in the correct order of operations for mathematics and for various programming languages.
$endgroup$
– naslundx
Apr 24 '14 at 16:57
$begingroup$
For cases like this, where there is no ubiquitous convention, the denotation is ambiguous, so you should always use parentheses to disambiguate.
$endgroup$
– Bill Dubuque
Apr 24 '14 at 17:08
$begingroup$
I would but with a test question they don't alway do stuff like that.
$endgroup$
– traisjames
Apr 24 '14 at 17:42
1
$begingroup$
@naslundx: examples show the order for Google is not the same as for the other softwares tested. Which is order "for mathematics"? $amod bcdot c=amod (bcdot c)$ or $amod bcdot c=(amod b)cdot c$? Shouldn't this be the answer?
$endgroup$
– MattAllegro
Apr 24 '14 at 19:25