Why Divide does not give the same answer?Prefix operator with low precedencewhen is f@g not the same as f[g]?How to construct custom operators with precedence?
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Why Divide does not give the same answer?
Prefix operator with low precedencewhen is f@g not the same as f[g]?How to construct custom operators with precedence?
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$begingroup$
I saw a question on Facebook $8div2 (2 + 2)=?$
Consider these inputs.
Divide[8, 2 (2 + 2)]
1
$8div2 (2 + 2)$ using esc+div+esc
16
Why the results are different.
I also tried these entries.
8/2 (2 + 2)
16
8/(2 (2 + 2))
1
Precedence /@ Plus, Subtract, Times, Divide
310., 310., 400., 470.
precedence
asked 8 hours ago
Okkes DulgerciOkkes Dulgerci
5,9671 gold badge11 silver badges21 bronze badges
$endgroup$
add a comment |
$begingroup$
I saw a question on Facebook $8div2 (2 + 2)=?$
Consider these inputs.
Divide[8, 2 (2 + 2)]
1
$8div2 (2 + 2)$ using esc+div+esc
16
Why the results are different.
I also tried these entries.
8/2 (2 + 2)
16
8/(2 (2 + 2))
1
Precedence /@ Plus, Subtract, Times, Divide
310., 310., 400., 470.
precedence
asked 8 hours ago
Okkes DulgerciOkkes Dulgerci
5,9671 gold badge11 silver badges21 bronze badges
$endgroup$
$begingroup$
Try usingTrace[]to see how the evaluation is performed.
$endgroup$
– Anjan Kumar
8 hours ago
$begingroup$
Is Mathematica just following the usual order of operations: en.m.wikipedia.org/wiki/Order_of_operations ?
$endgroup$
– JimB
8 hours ago
$begingroup$
@OkkesDulgerci The arguments toDivideare treated as if you had parenthesis around each one. There is no spilling from denominator to numerator because of operator precedence like you can have when you are writing/as infix operator.
$endgroup$
– Thies Heidecke
8 hours ago
$begingroup$
Then question is how should enter to find out the question?
$endgroup$
– Okkes Dulgerci
8 hours ago
1
$begingroup$
The answer to the question should be "It is X if I follow these rules, and Y if I follow these rules." It would be nice if that is what the Facebook poster is after. But my bias is that it is not an attempt at enlightenment.
$endgroup$
– JimB
8 hours ago
add a comment |
$begingroup$
I saw a question on Facebook $8div2 (2 + 2)=?$
Consider these inputs.
Divide[8, 2 (2 + 2)]
1
$8div2 (2 + 2)$ using esc+div+esc
16
Why the results are different.
I also tried these entries.
8/2 (2 + 2)
16
8/(2 (2 + 2))
1
Precedence /@ Plus, Subtract, Times, Divide
310., 310., 400., 470.
precedence
asked 8 hours ago
Okkes DulgerciOkkes Dulgerci
5,9671 gold badge11 silver badges21 bronze badges
$endgroup$
I saw a question on Facebook $8div2 (2 + 2)=?$
Consider these inputs.
Divide[8, 2 (2 + 2)]
1
$8div2 (2 + 2)$ using esc+div+esc
16
Why the results are different.
I also tried these entries.
8/2 (2 + 2)
16
8/(2 (2 + 2))
1
Precedence /@ Plus, Subtract, Times, Divide
310., 310., 400., 470.
precedence
precedence
asked 8 hours ago
Okkes DulgerciOkkes Dulgerci
5,9671 gold badge11 silver badges21 bronze badges
asked 8 hours ago
Okkes DulgerciOkkes Dulgerci
5,9671 gold badge11 silver badges21 bronze badges
asked 8 hours ago
Okkes DulgerciOkkes Dulgerci
5,9671 gold badge11 silver badges21 bronze badges
asked 8 hours ago
Okkes DulgerciOkkes Dulgerci
5,9671 gold badge11 silver badges21 bronze badges
asked 8 hours ago
Okkes DulgerciOkkes Dulgerci
5,9671 gold badge11 silver badges21 bronze badges
5,9671 gold badge11 silver badges21 bronze badges
$begingroup$
Try usingTrace[]to see how the evaluation is performed.
$endgroup$
– Anjan Kumar
8 hours ago
$begingroup$
Is Mathematica just following the usual order of operations: en.m.wikipedia.org/wiki/Order_of_operations ?
$endgroup$
– JimB
8 hours ago
$begingroup$
@OkkesDulgerci The arguments toDivideare treated as if you had parenthesis around each one. There is no spilling from denominator to numerator because of operator precedence like you can have when you are writing/as infix operator.
$endgroup$
– Thies Heidecke
8 hours ago
$begingroup$
Then question is how should enter to find out the question?
$endgroup$
– Okkes Dulgerci
8 hours ago
1
$begingroup$
The answer to the question should be "It is X if I follow these rules, and Y if I follow these rules." It would be nice if that is what the Facebook poster is after. But my bias is that it is not an attempt at enlightenment.
$endgroup$
– JimB
8 hours ago
add a comment |
$begingroup$
Try usingTrace[]to see how the evaluation is performed.
$endgroup$
– Anjan Kumar
8 hours ago
$begingroup$
Is Mathematica just following the usual order of operations: en.m.wikipedia.org/wiki/Order_of_operations ?
$endgroup$
– JimB
8 hours ago
$begingroup$
@OkkesDulgerci The arguments toDivideare treated as if you had parenthesis around each one. There is no spilling from denominator to numerator because of operator precedence like you can have when you are writing/as infix operator.
$endgroup$
– Thies Heidecke
8 hours ago
$begingroup$
Then question is how should enter to find out the question?
$endgroup$
– Okkes Dulgerci
8 hours ago
1
$begingroup$
The answer to the question should be "It is X if I follow these rules, and Y if I follow these rules." It would be nice if that is what the Facebook poster is after. But my bias is that it is not an attempt at enlightenment.
$endgroup$
– JimB
8 hours ago
$begingroup$
Try using
Trace[] to see how the evaluation is performed.$endgroup$
– Anjan Kumar
8 hours ago
$begingroup$
Try using
Trace[] to see how the evaluation is performed.$endgroup$
– Anjan Kumar
8 hours ago
$begingroup$
Is Mathematica just following the usual order of operations: en.m.wikipedia.org/wiki/Order_of_operations ?
$endgroup$
– JimB
8 hours ago
$begingroup$
Is Mathematica just following the usual order of operations: en.m.wikipedia.org/wiki/Order_of_operations ?
$endgroup$
– JimB
8 hours ago
$begingroup$
@OkkesDulgerci The arguments to
Divide are treated as if you had parenthesis around each one. There is no spilling from denominator to numerator because of operator precedence like you can have when you are writing / as infix operator.$endgroup$
– Thies Heidecke
8 hours ago
$begingroup$
@OkkesDulgerci The arguments to
Divide are treated as if you had parenthesis around each one. There is no spilling from denominator to numerator because of operator precedence like you can have when you are writing / as infix operator.$endgroup$
– Thies Heidecke
8 hours ago
$begingroup$
Then question is how should enter to find out the question?
$endgroup$
– Okkes Dulgerci
8 hours ago
$begingroup$
Then question is how should enter to find out the question?
$endgroup$
– Okkes Dulgerci
8 hours ago
1
1
$begingroup$
The answer to the question should be "It is X if I follow these rules, and Y if I follow these rules." It would be nice if that is what the Facebook poster is after. But my bias is that it is not an attempt at enlightenment.
$endgroup$
– JimB
8 hours ago
$begingroup$
The answer to the question should be "It is X if I follow these rules, and Y if I follow these rules." It would be nice if that is what the Facebook poster is after. But my bias is that it is not an attempt at enlightenment.
$endgroup$
– JimB
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, since the precedence of Divide is higher than Times, you should expect to parse 8 ÷ 2(2+2) as:
Divide[8, 2] (2+2)
You can also verify this by entering the input into a cell and using Cell | Show Expression to see what the boxes look like. The rendered version:
8 ÷ 2 (2 + 2)
and the version after using Cell | Show Expression:
Cell[BoxData[
RowBox[
RowBox["8", "[Divide]", "2"],
RowBox["(",
RowBox["2", "+", "2"], ")"]]], "Input",
CellLabel->"In[353]:="]
The boxes show that 8 ÷ 2 (with the boxes RowBox["8" "[Divide]", "2"]) is being multiplied by 2 + 2 (with the boxes RowBox["2", "+", "2"]).
answered 7 hours ago
Carl WollCarl Woll
88.7k3 gold badges117 silver badges228 bronze badges
$endgroup$
1
$begingroup$
AlsoFullForm[Hold[8/2 (2 + 2)]]might be easier to understand than "show expression"
$endgroup$
– Gustavo Delfino
4 hours ago
$begingroup$
Thanks. This clears things a bit. But my main concern was, on help page forDivideit says you can use esc+div+esc for shorthand $div$ and I expected to get the same result as Divide[8, 2 (2 + 2)]=1 but it didn't.
$endgroup$
– Okkes Dulgerci
34 mins ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, since the precedence of Divide is higher than Times, you should expect to parse 8 ÷ 2(2+2) as:
Divide[8, 2] (2+2)
You can also verify this by entering the input into a cell and using Cell | Show Expression to see what the boxes look like. The rendered version:
8 ÷ 2 (2 + 2)
and the version after using Cell | Show Expression:
Cell[BoxData[
RowBox[
RowBox["8", "[Divide]", "2"],
RowBox["(",
RowBox["2", "+", "2"], ")"]]], "Input",
CellLabel->"In[353]:="]
The boxes show that 8 ÷ 2 (with the boxes RowBox["8" "[Divide]", "2"]) is being multiplied by 2 + 2 (with the boxes RowBox["2", "+", "2"]).
answered 7 hours ago
Carl WollCarl Woll
88.7k3 gold badges117 silver badges228 bronze badges
$endgroup$
1
$begingroup$
AlsoFullForm[Hold[8/2 (2 + 2)]]might be easier to understand than "show expression"
$endgroup$
– Gustavo Delfino
4 hours ago
$begingroup$
Thanks. This clears things a bit. But my main concern was, on help page forDivideit says you can use esc+div+esc for shorthand $div$ and I expected to get the same result as Divide[8, 2 (2 + 2)]=1 but it didn't.
$endgroup$
– Okkes Dulgerci
34 mins ago
add a comment |
$begingroup$
First, since the precedence of Divide is higher than Times, you should expect to parse 8 ÷ 2(2+2) as:
Divide[8, 2] (2+2)
You can also verify this by entering the input into a cell and using Cell | Show Expression to see what the boxes look like. The rendered version:
8 ÷ 2 (2 + 2)
and the version after using Cell | Show Expression:
Cell[BoxData[
RowBox[
RowBox["8", "[Divide]", "2"],
RowBox["(",
RowBox["2", "+", "2"], ")"]]], "Input",
CellLabel->"In[353]:="]
The boxes show that 8 ÷ 2 (with the boxes RowBox["8" "[Divide]", "2"]) is being multiplied by 2 + 2 (with the boxes RowBox["2", "+", "2"]).
answered 7 hours ago
Carl WollCarl Woll
88.7k3 gold badges117 silver badges228 bronze badges
$endgroup$
1
$begingroup$
AlsoFullForm[Hold[8/2 (2 + 2)]]might be easier to understand than "show expression"
$endgroup$
– Gustavo Delfino
4 hours ago
$begingroup$
Thanks. This clears things a bit. But my main concern was, on help page forDivideit says you can use esc+div+esc for shorthand $div$ and I expected to get the same result as Divide[8, 2 (2 + 2)]=1 but it didn't.
$endgroup$
– Okkes Dulgerci
34 mins ago
add a comment |
$begingroup$
First, since the precedence of Divide is higher than Times, you should expect to parse 8 ÷ 2(2+2) as:
Divide[8, 2] (2+2)
You can also verify this by entering the input into a cell and using Cell | Show Expression to see what the boxes look like. The rendered version:
8 ÷ 2 (2 + 2)
and the version after using Cell | Show Expression:
Cell[BoxData[
RowBox[
RowBox["8", "[Divide]", "2"],
RowBox["(",
RowBox["2", "+", "2"], ")"]]], "Input",
CellLabel->"In[353]:="]
The boxes show that 8 ÷ 2 (with the boxes RowBox["8" "[Divide]", "2"]) is being multiplied by 2 + 2 (with the boxes RowBox["2", "+", "2"]).
answered 7 hours ago
Carl WollCarl Woll
88.7k3 gold badges117 silver badges228 bronze badges
$endgroup$
First, since the precedence of Divide is higher than Times, you should expect to parse 8 ÷ 2(2+2) as:
Divide[8, 2] (2+2)
You can also verify this by entering the input into a cell and using Cell | Show Expression to see what the boxes look like. The rendered version:
8 ÷ 2 (2 + 2)
and the version after using Cell | Show Expression:
Cell[BoxData[
RowBox[
RowBox["8", "[Divide]", "2"],
RowBox["(",
RowBox["2", "+", "2"], ")"]]], "Input",
CellLabel->"In[353]:="]
The boxes show that 8 ÷ 2 (with the boxes RowBox["8" "[Divide]", "2"]) is being multiplied by 2 + 2 (with the boxes RowBox["2", "+", "2"]).
answered 7 hours ago
Carl WollCarl Woll
88.7k3 gold badges117 silver badges228 bronze badges
answered 7 hours ago
Carl WollCarl Woll
88.7k3 gold badges117 silver badges228 bronze badges
answered 7 hours ago
Carl WollCarl Woll
88.7k3 gold badges117 silver badges228 bronze badges
answered 7 hours ago
Carl WollCarl Woll
88.7k3 gold badges117 silver badges228 bronze badges
88.7k3 gold badges117 silver badges228 bronze badges
1
$begingroup$
AlsoFullForm[Hold[8/2 (2 + 2)]]might be easier to understand than "show expression"
$endgroup$
– Gustavo Delfino
4 hours ago
$begingroup$
Thanks. This clears things a bit. But my main concern was, on help page forDivideit says you can use esc+div+esc for shorthand $div$ and I expected to get the same result as Divide[8, 2 (2 + 2)]=1 but it didn't.
$endgroup$
– Okkes Dulgerci
34 mins ago
add a comment |
1
$begingroup$
AlsoFullForm[Hold[8/2 (2 + 2)]]might be easier to understand than "show expression"
$endgroup$
– Gustavo Delfino
4 hours ago
$begingroup$
Thanks. This clears things a bit. But my main concern was, on help page forDivideit says you can use esc+div+esc for shorthand $div$ and I expected to get the same result as Divide[8, 2 (2 + 2)]=1 but it didn't.
$endgroup$
– Okkes Dulgerci
34 mins ago
1
1
$begingroup$
Also
FullForm[Hold[8/2 (2 + 2)]] might be easier to understand than "show expression"$endgroup$
– Gustavo Delfino
4 hours ago
$begingroup$
Also
FullForm[Hold[8/2 (2 + 2)]] might be easier to understand than "show expression"$endgroup$
– Gustavo Delfino
4 hours ago
$begingroup$
Thanks. This clears things a bit. But my main concern was, on help page for
Divide it says you can use esc+div+esc for shorthand $div$ and I expected to get the same result as Divide[8, 2 (2 + 2)]=1 but it didn't.$endgroup$
– Okkes Dulgerci
34 mins ago
$begingroup$
Thanks. This clears things a bit. But my main concern was, on help page for
Divide it says you can use esc+div+esc for shorthand $div$ and I expected to get the same result as Divide[8, 2 (2 + 2)]=1 but it didn't.$endgroup$
– Okkes Dulgerci
34 mins ago
add a comment |
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$begingroup$
Try using
Trace[]to see how the evaluation is performed.$endgroup$
– Anjan Kumar
8 hours ago
$begingroup$
Is Mathematica just following the usual order of operations: en.m.wikipedia.org/wiki/Order_of_operations ?
$endgroup$
– JimB
8 hours ago
$begingroup$
@OkkesDulgerci The arguments to
Divideare treated as if you had parenthesis around each one. There is no spilling from denominator to numerator because of operator precedence like you can have when you are writing/as infix operator.$endgroup$
– Thies Heidecke
8 hours ago
$begingroup$
Then question is how should enter to find out the question?
$endgroup$
– Okkes Dulgerci
8 hours ago
1
$begingroup$
The answer to the question should be "It is X if I follow these rules, and Y if I follow these rules." It would be nice if that is what the Facebook poster is after. But my bias is that it is not an attempt at enlightenment.
$endgroup$
– JimB
8 hours ago