YAM(ath)T: What does that sign mean?

Aves

Lifer
Feb 7, 2001
12,232
30
101
I think it means they didn't have the required font on their computer! :p
 

imported_Pablo

Diamond Member
Jan 20, 2002
3,714
1
0
looks like one of the default characters that a font uses when it can't find the right character...

(i confuse myself...)
 

ndee

Lifer
Jul 18, 2000
12,680
1
0
Originally posted by: aves2k
I think it means they didn't have the required font on their computer! :p

That was my first thought but how can I be sure?
 

Tauren

Diamond Member
Jan 30, 2001
3,880
1
0
It just couldn't find the appropriate font so it inserted a space holder.
 

chuckywang

Lifer
Jan 12, 2004
20,133
1
0
Yeah, it's definitely not rendered right. Hmm...my guess at what it is supposed to be is Z+, the positive integers.
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
81
Originally posted by: ndee
Originally posted by: aves2k
I think it means they didn't have the required font on their computer! :p

That was my first thought but how can I be sure?

For that equation to make any sense that symbol should either be Z R or N with a double linethrough it. Such as on these pages:
http://mathworld.wolfram.com/RealNumber.html
http://mathworld.wolfram.com/PositiveInteger.html


Edit: Oh, and there is no standard math operator that looks like that AFAIK.
 

ndee

Lifer
Jul 18, 2000
12,680
1
0
Originally posted by: chuckywang
Yeah, it's definitely not rendered right. Hmm...my guess at what it is supposed to be is Z+, the positive integers.

after 10 pages, I think I figured it out it's all the "natural" numbers, that's what we call it in German.
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: Kyteland
Originally posted by: ndee
Originally posted by: aves2k
I think it means they didn't have the required font on their computer! :p

That was my first thought but how can I be sure?

For that equation to make any sense that symbol should either be Z R or N with a double linethrough it. Such as on these pages:
http://mathworld.wolfram.com/RealNumber.html
http://mathworld.wolfram.com/PositiveInteger.html


Edit: Oh, and there is no standard math operator that looks like that AFAIK.

Actually there is. The box (actually box^2) is d'Alembertian. It's sort of the 4-d analogue to the Laplacian.

Laplacian = upside down triangle^2 = (d/dx)^2 + (d/dy)^2 + (d/dz)^2
d'Alembertian = box^2 = -(d/dt)^2 + (d/dx)^2 + (d/dy)^2 + (d/dz)^2

Mathworld
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
81
Originally posted by: silverpig
Actually there is. The box (actually box^2) is d'Alembertian. It's sort of the 4-d analogue to the Laplacian.

Laplacian = upside down triangle^2 = (d/dx)^2 + (d/dy)^2 + (d/dz)^2
d'Alembertian = box^2 = -(d/dt)^2 + (d/dx)^2 + (d/dy)^2 + (d/dz)^2

Mathworld
That's more of a square than a rectangle....

</orneryBastard>

;)