Operations that Maintain/Don't Maintain Inequality

[madah12]

Homework Statement

then what are the operations that maintain the Inequality and what are the operations that don't?

Homework Equations

The Attempt at a Solution

clearly addition and subtraction maintains it ,and so does multiplication and division by any number other than zero. also taking any nonzero exponent except incase numbers a and -a
cause a !=-a but a^even = (-a)^even and the zeroes but what about the logarithms?
2 =! -2 but can we say that ln(-2) =! ln(2) I mean since there is no such thing as ln(-2) can we not equalize something we don't know? and what are the other operations that I forgot to mention that maintain or don't maintain the inequality

 

[LeonhardEuler]

clearly addition and subtraction maintains it ,and so does multiplication and division by any number other than zero.

Are you sure? We know
2 > 1
Now multiply or divide both sides by -1. Does the inequality hold still?

 

[madah12]

i am not talking about > or < I am talking about =!
2 =! 1
-2 =! -1
yes that holds

 

[LeonhardEuler]

Ok, misunderstood. Do you know about 1-to-1 functions?

 

[madah12]

no I don't even know how to correctly read that...

 

[LeonhardEuler]

Oh, well you can look here http://en.wikipedia.org/wiki/One-to-one. It has a lot of relevance to the problem, I think. Think about functions that are not 1-to-1 like sin and cos and x^2 compared to ones that are like e^x, x^3 and arctan

 

[madah12]

so you are implying that if f is Injective then if a =! b , f(a) =! f(b) but if it isn't injective then there might be be numbers a,b where a =! b but f(a)=f(b)?

 

[LeonhardEuler]

You got it except that it isn't just that there might be numbers a,b where a =! b but f(a)=f(b) when f isn't injective, there definitely are.

 

[madah12]

oh I see thanks but can you help me in the logarithm part of whether we can say than ln(2) IS not equal to ln(-2) even if we don't know what ln(-2) is?

 

[LeonhardEuler]

You need to add another requirement that the domain of the function needs to include all the allowed numbers.

 

[Char. Limit]

$$ln(-2) = ln(-1) + ln(2) = (2k+1) i \pi + ln(2), k\epsilon \mathbb{N}$$

This is, of course, if you allow complex numbers.

 

[LeonhardEuler]

Although that has it's own problems and ln(x) is usually restricted to the real numbers. In the complex case every number has infinitely many logarithms, and the "principle value" of the logarithm is usually denoted
$$\text{Log}(z)$$
with a capitol L.