When we say that two algebraic expressions, [tex]f(x_1,x_2,\cdots

  • Thread starter dalcde
  • Start date
  • #1
166
0

Main Question or Discussion Point

When we say that two algebraic expressions, [tex]f(x_1,x_2,\cdots x_n)[/tex], [tex]g(x_1,x_2,\cdots x_n)[/tex] are identical, or
[tex]f(x_1,x_2,\cdots x_n)\equiv g(x_1,x_2,\cdots x_n)[/tex], we mean (according to my textbook) that [tex]\forall x_1,x_2,\cdots x_n: f(x_1,x_2,\cdots x_n)=g(x_1,x_2,\cdots x_n)[/tex]
However, does that mean that
[tex]x^2+x\equiv0[/tex]
in [tex]Z^2[/tex]?

If not, does it mean that
[tex]f(x_1,x_2,\cdots x_n)\equiv g(x_1,x_2,\cdots x_n)\Leftrightarrow\forall\phi: \phi(f(x_1,x_2,\cdots x_n))=\phi(g(x_1,x_2,\cdots x_n))[/tex]
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
132


When we say that two algebraic expressions, [tex]f(x_1,x_2,\cdots x_n)[/tex], [tex]g(x_1,x_2,\cdots x_n)[/tex] are identical, or
[tex]f(x_1,x_2,\cdots x_n)\equiv g(x_1,x_2,\cdots x_n)[/tex], we mean (according to my textbook) that [tex]\forall x_1,x_2,\cdots x_n: f(x_1,x_2,\cdots x_n)=g(x_1,x_2,\cdots x_n)[/tex]
However, does that mean that
[tex]x^2+x\equiv0[/tex]
in [tex]Z^2[/tex]?

If not, does it mean that
[tex]f(x_1,x_2,\cdots x_n)\equiv g(x_1,x_2,\cdots x_n)\Leftrightarrow\forall\phi: \phi(f(x_1,x_2,\cdots x_n))=\phi(g(x_1,x_2,\cdots x_n))[/tex]
Are you talking about number theoretic congruences? If so you need a modulus to be defined if this is the case.

Assuming this is the case I imagine that under congruences the strength of your phi relation may not hold in general.

Take for example mod 10. 3 Mod 10 = 3 and phi(3) = 2. 13 Mod 10 = 3 but phi(13) = 12 (since both are primes).

Is this what you mean? Maybe it isn't though, so some feedback would be appreciated.
 
  • #3
166
0


I'm talking about the identities in the simple high-school sense, in case that's what you are asking.

My [tex]\phi[/tex]s are propositional functions, which return a true or false answer. For example, [tex]\phi(f(x))[/tex] might be the statement "f(x) is a polynomial with degree 2", in which it is true for [tex]x^2+x[/tex] but not 0. I got that notation from Principia Mathematica (by the way, I surrendered after reading less than half a volume - it was too tough). Please tell me if my notation is too outdated.
 
  • #4
chiro
Science Advisor
4,790
132


I'm talking about the identities in the simple high-school sense, in case that's what you are asking.

My [tex]\phi[/tex]s are propositional functions, which return a true or false answer. For example, [tex]\phi(f(x))[/tex] might be the statement "f(x) is a polynomial with degree 2", in which it is true for [tex]x^2+x[/tex] but not 0. I got that notation from Principia Mathematica (by the way, I surrendered after reading less than half a volume - it was too tough). Please tell me if my notation is too outdated.
Ohh ok no worries. I'm doing a cryptography course at the moment so phi's to me are the Euler totient function and I put your post in the context of number theory due to your congruence symbols.

I'm not really qualified to answer your question since I'm unfamiliar with the notation. The principia is an old document though. But if liebniz notation has survived this long, maybe the notation in the principia has a chance :)
 
  • #5
AlephZero
Science Advisor
Homework Helper
6,994
291


However, does that mean that
[tex]x^2+x\equiv0[/tex]

in [tex]Z^2[/tex]?
Yes, by your textbook definition.

If not, does it mean that
[tex]f(x_1,x_2,\cdots x_n)\equiv g(x_1,x_2,\cdots x_n)\Leftrightarrow\forall\phi: \phi(f(x_1,x_2,\cdots x_n))=\phi(g(x_1,x_2,\cdots x_n))[/tex]
You need to understand the difference between a function [itex]\phi[/itex] whose arguments are the values of [itex]f[/itex] (i.e. all possible values, depending on the values of [itex]x_1,x_2,\cdots x_n[/itex]), and a function whose argument is the function [itex]f[/itex] itself.

IIRC Principia uses the different notations

[tex]\phi f[/tex] and

[tex]\hat\phi f[/tex]

when the difference is important.
 

Related Threads on When we say that two algebraic expressions, [tex]f(x_1,x_2,\cdots

Replies
9
Views
2K
Replies
5
Views
4K
Replies
9
Views
3K
Replies
7
Views
3K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
4
Views
468
Replies
4
Views
701
  • Last Post
Replies
7
Views
3K
Replies
9
Views
8K
Top