Maths Question

Originally posted by FabianFnas
I would love to see the six first rules of yours! 🙂

1) the additive inverse is unique

2) the additive identity is unique

3) (-1)*a = (-a)

4) (-a)*(-b) = a*b

5) 0*a = 0

6) the multiplicative inverse is unique

Originally posted by FabianFnas
I would love to see the six first rules of yours! 🙂

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 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.

Originally posted by mtthw
Axioms of the real numbers! As my Analysis I course taught it.

Great! I thought you was joking, but now I lift my hat for you!

Originally posted by Dejection
How about 0^0?

x^0 = 1
0^x = 1

So what is 0^0?

0^0 is defined as being equal to:-
Limit x^x when x -->0 [i.e. x tends to zero]
This limit works out to 1.
Hence 0^0 = 1.

Originally posted by Fat Lady
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!

0^0 = 1 not because of bionomial theorem;
But exactly because Limit of x^x as x-->0 works out to 1.

Originally posted by CoolPlayer
0^0 = 1 not because of bionomial theorem;
But exactly because Limit of x^x as x-->0 works out to 1.

But the limit of 0^x doesn't equal 1 as x approaches zero.

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.

Originally posted by Fat Lady
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.

Surely the definition of powers (and therefore 0^0) predates the theorems it might break?

Originally posted by Diapason
If you keep asking questions like this then I'll need to send you to l'Hopital...

😀

Originally posted by mtthw
Actually, it may have been no. 8: 0 != 1 may have been a separate axiom. Anyway, there are then four or five more.

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.

Originally posted by adam warlock
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.

Not practical, but entirely self-consistent. Which is why you need it - "the non-triviality axiom".

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.

Originally posted by twilight2007
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.

Also, by definition, 0^0=0/0, so that too is indeterminate.

I don't understand. What definition?

Originally posted by mtthw
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.

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.

My mother taught me two things; "beware of ladies of a certain virtue" and "never divide by zero"!
-a lecturer

Originally posted by genius
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.

I know. That's why I said "axioms of the real numbers".

Originally posted by mtthw
I know. That's why I said "axioms of the real numbers".

I know, but you initially said, "It was rule no.7 the way we were taught", but that is meerly for the real numbers...