I've seen a notebook with the following:

 Solve[{eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9, eq10, eq11, eq12},
       {hbbbf, hbcf, hbbcr, hbcc, hbvbf, hbbc, hsbbf, hssd, hsvbf, hsf, hsbr, hsc}
][[1]] // N

I searched the Mathemtatica help documentation for // and it seems to be a postfix operator. On the other hand N seems to be a way to specify the precision of a numerical value.

What does the [[1]] // N mean?



2 Answers 2


Actually, // is not a postfix operator, itself; it would be considered an infix operator, akin to + or -. It operates by turning x // f into f[x]. You can string several of them together, e.g. x // f // g which is equivalent to g[f[x]], and think of them as successive transformations. This can be very useful for crafting complex transformations, but it does make for very opaque code. Nowadays, I tend to create many functions and layer them to create the effects which makes my code a bit easier to follow. Similarly, there is the prefix form f @ x which is operationally equivalent to x // f.

As pointed out in Michael's answer, the entire left hand side (LHS) must be considered as the argument to the right hand side (RHS) which usually works. There are cases where it will not, but that depends on the precedence of the operators used on the LHS. For most cases, there is nothing to worry about as // has a relatively low precedence, but consider the following

 f = x // g

The Set operator (=) has a lower precedence than //, so the above is equivalent to

f = g[x]


Additionally, mixing operators with higher precedence can cause some confusion. For instance, the following

m // p + q // n

is interpreted as

n[ (p + q)[m] ]

which is not likely what you would want, so parentheses should be used, such as

(m // p) + q // n
(* n[q + p[m]] *)


(m // p) + (q // n)
(* n[q] + p[m] *)

depending on what you want.

It is interesting to note that the function operator (&) which turns the preceding into a pure function has a low precedence relative to most operators, but a higher precedence than //. So, this

m // p + q& // n

is interpreted as

n[ Function[ p + q ][m] ]


n[ p + q ]

if neither p nor q are slots.


Part ([[i]]), on the other hand, has a high precedence, so it will take effect prior to //, and it can be considered a postfix operator, also. In other words, it effects the output from FullSimplify prior to N being applied. Rewriting the code as it is interpreted, you get

N[ Part[ FullSimplify[ Solve[ ... ] ], 1 ] ]

But, I am partial to using the prefix form here

N @ FullSimplify[ ... ][[1]]

or, possibly more clearly

N @ First @ FullSimplify[ ... ]

as the intent of the code is clarified a bit by letting the user know they will be getting a numerical result up front. The shift to using First instead of Part[..., 1] only further clarifies what is occurring. The downside to this construct, as opposed to the postfix form, is it can be more difficult to add additional transformations while keeping them straight in your mind. For the so inclined, here is the full postfix form

FullSimplify[ ... ] // First // N

or, how I would write it

Solve[ ... ] // FullSimplify // First // N

Lastly, while N can be used to specify numerical precision, I primarily use it to convert from analytic numbers to numerical numbers. For instance, Mathematica is perfectly happy to work with 1/3, but sometimes I need a numerical answer. So, I use N to perform the conversion.

  • $\begingroup$ I see you decided to make something of this question. +1 :-) $\endgroup$
    – Mr.Wizard
    Jan 6, 2013 at 5:41
  • $\begingroup$ @Mr.Wizard every once in a while, you can turn a so-so question into a teaching moment, and expose and clarify misconceptions. Besides, that link to the operator precedence list is not as easy to find as I'd like (I know what to look for, so I can find it), so I like to have it referenced as often as possible. $\endgroup$
    – rcollyer
    Jan 6, 2013 at 6:52
  • $\begingroup$ Exactly. I'm glad to see you do it. $\endgroup$
    – Mr.Wizard
    Jan 6, 2013 at 6:53
  • 2
    $\begingroup$ @Mr.Wizard my goal: don't have mma.se become like the c++ tag on SO. Basic misconceptions can be answered, and should. Sometimes a nudge is all that is necessary, and that is what comments are for. But, occasionally they need a thorough analysis, otherwise they may persist, and this one caused enough mental friction with me, that I decided to knock it out of the park. $\endgroup$
    – rcollyer
    Jan 6, 2013 at 7:32
  • $\begingroup$ @badb, just let me add a link to the official operator precedence documentation. And as for a quick guide this one. $\endgroup$
    – carlosayam
    Jan 6, 2013 at 12:13
  1. [[1]] means part 1, the first element of the expression returned by FullSimplify. FullSimplify will return the simplified list of solutions (found by Solve) of the twelve equations.

  2. // N converts exact numbers to (approximate) Real numbers.

The right way to parse the expression is (FullSimplify[...][[1]]) // N.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.