20 Sep 07
3 edits
Originally posted by FabianFnasAxioms of the real numbers! As my Analysis I course taught it.
I would love to see the six first rules of yours! 🙂
1. addition and multiplication commutative
2. addition and multiplication associative
3. multiplication distributive over addition
4. existence of additive identity (0)
5. existence of multiplicative identity (1, != 0)
6. existence of additive inverse
7. existence of multiplicative inverse (for x != 0)
It's No. 7 that defines division, and specifically doesn't define it for 0. So you can't divide by zero.
Actually, it may have been no. 8: 0 != 1 may have been a separate axiom. Anyway, there are then four or five more.
20 Sep 07
Originally posted by Fat Lady0^0 = 1 not because of bionomial theorem;
I assume you meant:
x^0 = 1
0^x = 0 (or if you like e^x tends to 0 as e tends to 0).
So what is 0^0.
The answer is this is actually defined as being equal to 1, mainly so that the binomial theorem works!
But exactly because Limit of x^x as x-->0 works out to 1.
20 Sep 07
Originally posted by CoolPlayerBut the limit of 0^x doesn't equal 1 as x approaches zero.
0^0 = 1 not because of bionomial theorem;
But exactly because Limit of x^x as x-->0 works out to 1.
You are incorrect in stating that 0^0 = 1 because x^x tends to 1 as x tends to zero. 0^0 is taken to be 1 because it breaks less theorems if we say it is 1 as opposed to zero or indeterminate.
20 Sep 07
Originally posted by Fat LadySurely the definition of powers (and therefore 0^0) predates the theorems it might break?
You are incorrect in stating that 0^0 = 1 because x^x tends to 1 as x tends to zero. 0^0 is taken to be 1 because it breaks less theorems if we say it is 1 as opposed to zero or indeterminate.
23 Sep 07
Originally posted by mtthwAs I learned it it wasn't truly an axiom it was more of a necessary consequence. If you didn't request for them to be different the necessary condition of all the axioms was that all real numbers were equal. That is to say that only one real number existed. Not a very practical number system.
Actually, it may have been no. 8: 0 != 1 may have been a separate axiom. Anyway, there are then four or five more.
23 Sep 07
1 edit
Originally posted by adam warlockNot practical, but entirely self-consistent. Which is why you need it - "the non-triviality axiom".
As I learned it it wasn't truly an axiom it was more of a necessary consequence. If you didn't request for them to be different the necessary condition of all the axioms was that all real numbers were equal. That is to say that only one real number existed. Not a very practical number system.
03 Oct 07
0/0 is defined as an "indeterminate form". The quotation marks signify that I'm not sure if this is the most accurate term.
For example: a/0:
If a is a positive real number, 1/0 equals positive infinity.
If a is a negative real number, a/0 equals negative infinity.
If a=0, then a/0 is undefined.
Also, by definition, 0^0=0/0, so that too is indeterminate.
People who have studied algebra may know that you can solve for one variable in a rational expression if that variable is the only variable in the numerator. For example:
x/(abcdefghijklmnopqrstuvwyz) can equal 0 (x=0) as long as {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,y,z =/= 0}.
However, (x^2-2x+1)/(x-1)=0 does not have a solution, because the numerator (x^2-2x+1) only equals 0 when x=1, and if x=1, we obtain the indeterminate form 0/0, which doesn't equal 0 or 1. So, be careful when solving rational expressions and make sure that the denominator doesn't equal 0.
04 Oct 07
Originally posted by twilight2007Also, by definition, 0^0=0/0, so that too is indeterminate.
0/0 is defined as an "indeterminate form". The quotation marks signify that I'm not sure if this is the most accurate term.
For example: a/0:
If a is a positive real number, 1/0 equals positive infinity.
If a is a negative real number, a/0 equals negative infinity.
If a=0, then a/0 is undefined.
Also, by definition, 0^0=0/0, so that too is inde ...[text shortened]... reful when solving rational expressions and make sure that the denominator doesn't equal 0.
I don't understand. What definition?
07 Oct 07
Originally posted by mtthwThose are meerly for the Real numbers, not the entirety of mathematics. It is never possible to divide my 0, but commutativity is often false. For instance, the Quaternion group? i.j=-j.i, etc.
Axioms of the real numbers! As my Analysis I course taught it.
1. addition and multiplication commutative
2. addition and multiplication associative
3. multiplication distributive over addition
4. existence of additive identity (0)
5. existence of multiplicative identity (1, != 0)
6. existence of additive inverse
7. existence of multiplicative invers ...[text shortened]... been no. 8: 0 != 1 may have been a separate axiom. Anyway, there are then four or five more.
My mother taught me two things; "beware of ladies of a certain virtue" and "never divide by zero"!
-a lecturer
07 Oct 07
Originally posted by geniusI know. That's why I said "axioms of the real numbers".
Those are meerly for the Real numbers, not the entirety of mathematics. It is never possible to divide my 0, but commutativity is often false. For instance, the Quaternion group? i.j=-j.i, etc.