Men! I have a big problem here oh....

I find it difficult rendering number of elements in a set. For example:[math]n(A)[/math]

But that's not what I want to write. I want to have this shown in latex: ©(P\cap{Q})'

I wrote it like this:[math](c)(P\cap{Q})'[/math] and am having latex error. I have gone through the whole library of latex symbols;no help is sight. I need help please.

[math]\vert A \vert[/math], [math]\textnormal{n}(A)[/math], [math]\textnormal{card}(A)[/math], [math]\# A[/math], [math]\bar{\bar{A}}[/math]

 

Any of these help?

 

I am not sure of any other notation in common use.

[math]\vert A \vert[/math], [math]\textnormal{n}(A)[/math], [math]\textnormal{card}(A)[/math], [math]\# A[/math], [math]\bar{\bar{A}}[/math].

Mind you, this is what I actually want to write.

Edited August 20, 201411 yr by Chikis

per the actual thread the problem is the use of (c) - this is common word processor short form for the copyright symbol © - you see it has changed in your OP. The LaTex renderer obviously knows this and goes to look for the symbol - but for some reason (probably because it is legal symbol rather than mathematical) the LaTex we use does not include it in the font; therefore it returns an error. if you want a c in brackets in Latex you can just add an extra white space

[latex](c )[/latex]

ie

[math](c )[/math]

 

edit

 

although I note the noparse tag has stopped working again

 

ie

[_math](c )[_/math]

but removing underscores

[math]( C )(P \cap Q)[/math]

 

Something like the above?

[math]( C )(P \cap Q)[/math]Something like the above?

Not untill when it becomes

[math]( c )(P \cap Q)'[/math]

that c is a small letter. Is just like a kind of numbering the problems. We have problems a, b and c.

Edited August 21, 201411 yr by Chikis

[math]( a ) \hspace{10pt} E = m c^{2}[/math]

[math]( b ) \hspace{10pt} \textnormal{card}(P \cap Q)' = 78[/math]

[math]( c ) \hspace{10pt} y = mx +c[/math]

 

Something more like the above...

Never mind. This [math]( c )(P \cap Q)'[/math] has sufficed everything.