I can't be sure whether this is a bug or a programing choice but...

When you have an equation such as the one below...

a:= 1 + 2 + 3

and you want to change it to...

a:= 1 + (2 + 3)/4

you would like to be able to insert the curse on the right and then extend it under the 2 and 3 and then divide that sum by 4. However on my computer the curser, (using the arrow keys) skips over 2 and 3 and under-scores all three numbers, 1,2 and 3. When I expand the curser from the left hand side it does what you would expect it to.

(Win7 64bit, version 0.89 (build 3856)

Again maybe this is intentional but I can't see the reason one would want this behavior.

Dave

Radovan , as a result of my systematic study of curser behaviour in prep of my tutorial, I’ve come up with a second more general way of accomplishing the task.

Here is a ruff draft of what I’ve written for the tutorial. I’ll be dealing with screen shots only as a final step before posting.
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At his point we will discuss the systematic behaviour of the curser. Enter the equation, 1+2 +3 + 4 = and slowly move the curser back and forth studying its behaviour.
Below is a summary of the different curser positions as you move from left to right.

|1 + 2 + 3 + 4 = 10
|1 + 2 + 3 + 4 = 10
|1 + 2 + 3 + 4 = 10
|1 + 2 + 3 + 4 = 10
|1 + 2 + 3 + 4 = 10
1| + 2 + 3 + 4 = 10
1 + |2 + 3 + 4 = 10
1 + 2| + 3 + 4 = 10
1 + 2 + |3 + 4 = 10
1 + 2 + 3| + 4 = 10
1 + 2 + 3 + |4 = 10
1 + 2 + 3 + 4| = 10

The important thing to note here is that the only time the curser spans a range of elements is when that range begins at the far left.
We need to keep this in mind when we go to perform tasks. For example, say we wish to divide 3+4 by 5 in the above equation, since we can not extend the range of the curser under 3+4 we will instead use the fact that we can extend the range of the curser under 1+2.
Below are the required steps to divide 3+4 by 5.

|1 + 2 + 3 + 4 = 10 . . : Position the curser under 1+2
■ + 3 + 4 = 10 . . . . . . . . : Cut it onto the clipboard (Ctrl+X)
3 + 4 = 10 . . . . . . . . . . : Backspace over the addition operator and placeholder.
(3 + 4)/5 = 10 . . . . . . . . : Divide 3+4 by 5
+ (3 + 4)/5 = 10 . . . . . . . : Insert the addition operator
b + (3 + 4)/5 = 10 . . . . . . : Insert a temporary dummy value like b or any number
1 + 2 + (3 + 4)/5 = 10 . . . . : Paste the sum 1+2 back in, replacing the dummy value.

The dummy value was used in this instance because the program only allows you to paste a general expression like 1+2 into the location of an already existing text segment or text segment placeholder.

This method will of course work with any general expression; we used simple digits here for clarity purposes.

As it happens the curser behaviour described above occurs between any two brackets. This means that if we had the expression like the one below....

x^2 + ln(30)*(1 + 2 + 3 + 4) = xxx

the steps would have been exactly the same.

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Dave