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I've tracked a bug in my code down to the problem of adding two numbers together, with the left argument having machine precision, e.g by 3` . The issue is, if there is no space between the two numbers, then the addition acts on the precision of the number and not the number itself. For example,

3`+2 produces 3.0 instead of 5..

additionally, if we specify a numerical value for the precision, we get the correct result with specified precision:

3`2+2 produces 5.00

There is no space between the two arguments and the addition sign. If we add spaces:

3`+ 2 produces 5.

3` + 2 produces 5.

This possible bug caused me a big headache in my code. Is there a way to globally remove this from the notebook so that I get the expected result from, e.g. 3`+2=5.? I couldnt find anything in the documentation regarding this issue.

Some possibly relevant information:

My mathematica version is Version 13.0.1 for Linux x86 (64-bit)

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    $\begingroup$ For information about operator/input precedence see this answer. Note that Precedence[Precision]==670. and Precedence[Plus]==310., so an ambiguously entered input expression gives precedence to the backtick precedence rather than the +. $\endgroup$
    – evanb
    Jul 1 at 8:38
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    $\begingroup$ See reference.wolfram.com/language/tutorial/InputSyntax.html#7977 for the definition of the input number`s where s may be a decimal real number or integer. $\endgroup$
    – Michael E2
    Jul 1 at 16:57

2 Answers 2

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This is a well-known problem of parsing of ambiguous input in Mathematica.

If you enter 3`+2, it is correctly parsed as arbitrary precision number 3 with precision 2:

3`+2 // InputForm
3.`2.

If you include a space between the backtick and the plus sign, it will be interpreted as an addition of machine precision number 3 with exact number 2:

3` + 2
5.

Alternatively, you can specify that a number is a MachinePrecision number just by placing the dot right after it (in this case the presence or absence of the spaces between the dot and the plus sign does not matter):

3. + 2
5.

Is there a way to globally remove this from the notebook so that I get the expected result from, e.g. 3`+2=5.?

If I type in the "Find and Replace" dialog `+ (without a space) in the "Find:" field, and ` + (with the space character between the backtick and the plus sign) in the "Replace with:" field, I'm able to replace all the occurences, and obtan the result you wish (I'm on Mathematica 13.1.0).

screenshot

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Here I have a different take from @AlexeyPopkov (+1), as I think taking care of the number of spaces in an expression is not a desirable programming strategy for Mathematica, so our analysis is equivalent but our solutions or advice are different.

Analysis

The key point was made by @evanb, your expression

3`+2

Has two operators competing, SetPrecision and Plus and

Precedence[SetPrecision]>Precedence[Plus]
(* True *)

So it gets interpreted as

SetPrecision[3,Plus[2]]

and then

SetPrecision[3,2]

as Plus[2]==2, so you get a low precision 3 as output. On the other hand, with a space

3` +2

is unambiguous, the space breaks the ambiguity, no need to check precedence, and therefore interpreted as

Plus[SetPrecision[3,MachinePrecision],2]

which is 5. with MachinePrecision.

Solution

I would suggest avoiding potentially ambiguous and hard-to-debug expressions like

3`+2 

and instead, use FullForm syntax to input your code, as such is the most unambiguous way to get the behaviour you expect.

Plus[SetPrecision[3,MachinePrecision],2]

Taking care of the number of spaces in an expression is not a desirable programming strategy for Mathematica.

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    $\begingroup$ I should stress that $MachinePrecision and MachinePrecision are completely diffrent things. The first forces the arbitrary precision arithmetics, while the second is the actual machine precision number. You can see with Precision[3`] returning MachinePrecision that it is a MachinePrecision number, not an arbitrary precision. But Precision[SetPrecision[3, $MachinePrecision]] returns 15.954589770191003 what means that with SetPrecision[3, $MachinePrecision] you get an arbitrary precision number. $\endgroup$ Jul 1 at 15:58
  • $\begingroup$ @AlexeyPopkov thanks for the comment, I amended my answer. $\endgroup$
    – rhermans
    Jul 1 at 16:06
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    $\begingroup$ Further, addition of any number to a machine precision number gives a machine precision number. $\endgroup$ Jul 1 at 16:15
  • $\begingroup$ Corrected again @AlexeyPopkov, thanks $\endgroup$
    – rhermans
    Jul 1 at 16:20

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