is multiplication a repeating addition? and few other q's

[mohammad muazzam]

I have a few questions:
all of the alphabet characters are the questions (quite a lot). If you could help me with just a few of them, maybe i could figure the rest of them myself

let me explain one of the question.
[tex]4^{2}[/tex] = 4 [tex]\cdot[/tex] 4 = 4 + 4 + 4 + 4 = 16
exponent tells us how many times to multiply a number by itself. in this case, 2 times, so it would be 4 [tex]\cdot[/tex] 4. multiplication is just repeating addition, that's why it's 4 + 4 + 4 + 4. (this is what i understand)
but it gets confusing when the base is less than 1.
([tex]\frac{1}{2}[/tex]) [tex]^{2}[/tex] = [tex]\frac{1}{2}[/tex] [tex]\cdot[/tex] [tex]\frac{1}{2}[/tex] = ? = [tex]\frac{1}{4}[/tex]
x + y = [tex]\frac{1}{4}[/tex]
what would x and y be?
this is just one of the question that has been bothering my mind for the past few days.
(a) [tex]0.1^{0.1}[/tex] = 0.794...
(b) [tex]0.5^{0.5}[/tex] = 0.707...
how can equation (a) has a higher value than equation (b)?
thanks for the help

[HallsofIvy]

Your problem is in thinking that "multiplication is repeated addition". That is true only when one of the multiplicands is a positive integer.

Similarly "exponentiation is repeated multiplication" is only true if the exponent is a positive integer. IF n and m are positive integers then [tex]a^{n+m}= (a^n)(a^m)[/tex] and [tex](a^m)^n= a^{mn}[/tex]. Those are nice properties so while we are allowed to define "a to other powers" any way we want, we would like those to be true for m and n not positive integers.

For example, how should we define a to the 0 power? Well, for n any positive integer n+ 0= n. So we want [tex]a^n= a^{n+0}= a^n a^0[/tex]. For a non-zero, we can divide both sides by [tex]a^n[/tex] to get [tex]a^0= 1[/tex] so we define [tex]a^0= 1[/tex] and define it only for a non-zero.

n+ (-n)= 0 so we want [tex]1= a^0= a^{n- n}= a^n(a^{-n})[/tex] so we define [tex]a^{-n}= \frac{1}{a^n}[/tex].

n/n= n(1/n)= 1 so [tex](a^{1/n})^n= a^{n/n}= a[/tex] so we define [tex]a^{1/n}= \sqrt[n]{a}[/tex].

Of course then for any rational number m/n, [tex]a^{m/n}= (a^{1/n})^m= (\sqrt[n]{a})^m[/tex] as well as [tex]\sqrt[n]{a^m}[/tex].

Finally, the set of all real numbers cannot be defined "algebraically"- they have to be defined "analytically"- using a limit process. For example [tex]\pi[/tex] is irrational and is equal to the infinite decimal 3.1415926... which is the same as saying that [tex]\pi[/tex] is the limit of the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, 3.1415926... and we define [tex]a^{\pi}[/tex] as the limit of the sequence [tex]a^3[/tex], [tex]a^{3.1}[/tex], [tex]a^{3.14}[/tex], etc. (All of those are "terminating decimals" so rational numbers so the exponentials are already defined.)

It is, however, true that for a< b, [tex]a^a< b^b[/tex]. I'm not sure why you would be puzzled by that.

[HallsofIvy]

I realize that I misread your last question. IF 1< a< b then [tex]a^a< b^n[/tex] but your last question has both a and b less than 1 which reverses the situation.