# If a =! b

## Homework Statement

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

## 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 dont maintain the inequality

LeonhardEuler
Gold Member
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?

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

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

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

LeonhardEuler
Gold Member
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

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
Gold Member
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.

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 dont know what ln(-2) is?

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

Char. Limit
Gold Member
$$ln(-2) = ln(-1) + ln(2) = (2k+1) i \pi + ln(2), k\epsilon \mathbb{N}$$

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

LeonhardEuler
Gold Member
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.