Proof of a^xy=a^x+a^y

[georg gill]

This is the proof

http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html [Broken]

I wonder what they do when they describe monotnicity for integers. Why is

$$x^p<y<p$$ when x<y and p is a positive integer? What if p was really large?


and further down in the part homogeneity for rational exponents is there a typo there? Should not

$$(v_1 v_2)^n > (xy)^m$$

and not

$$(v_1 v_2)^n < (xy)^m$$

or?

 

[I like Serena]

Hi georg gill!

This is the proof

http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html [Broken]

I wonder what they do when they describe monotnicity for integers. Why is

$$x^p<y<p$$ when x<y and p is a positive integer? What if p was really large?

It's a typo. It should be:
$$x^p<y^p$$

and further down in the part homogeneity for rational exponents is there a typo there? Should not

$$(v_1 v_2)^n > (xy)^m$$

and not

$$(v_1 v_2)^n < (xy)^m$$

or?

Yep. You found another typo.

Good that you understand the material!

 

[HallsofIvy]

As to the title of this thread, however, they are proving a^(x+y)= a^xa^y and a^(xy)= (a^x)^y= (a^y)^x, NOT "a^(xy)= a^x+ a^y" which is simply not true.

 

[georg gill]

As to the title of this thread, however, they are proving a^(x+y)= a^xa^y and a^(xy)= (a^x)^y= (a^y)^x, NOT "a^(xy)= a^x+ a^y" which is simply not true.

sorry about the heading.


but the part


[tex] a^(xy)= (a^x)^y= (a^y)^x[\tex] (c)

I did not know this was a proof for (c). that was a proof I needed much. Very cool! I just have to work through it later and see if I get it:)

 

[I like Serena]

Which step do you not get exactly?

(TBH, I find it a bit troublesome to work through those web links. Perhaps you can copy the step that's giving you difficulty?)

 

[georg gill]

I guess this is my explenation for what I did get from the proof

http://bildr.no/view/1035536

But actually i wanted to prove that

$$\sqrt[n]{c^m}\sqrt[n]{d^m}=c^{m/n}d^{m/n}$$ (I)

dealing with integers when working with powers and roots are much simpler because it is easier to think about a number multiplied by itself n times and the nth roth of a number because it has to be a number one gets by doing that. If one could prove (I) then one could go from whole integers for n and m on the left side to a fraction on the right side and by that one would always see that one could first find a power and then a root even (or first root then power, to switch order of theese operations is provable) though it is written as the right side of (I)

I also did manage from another proof to make this explanation (I) only a matter of showing that

$$a^{\frac{1}{n}}a^{\frac{1}{n}}=a^{\frac{2}{n}}$$ (vv)

because

http://www.viewdocsonline.com/document/biwlgx

If we use (I)

$$\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$$

we could get (II):

$$\sqrt[n]{a^m}=\sqrt[n]{aa...a}=y$$

from (I)

$$\sqrt[n]{a^m}=\sqrt[n]{a}\sqrt[n]{a}...\sqrt[n]{a}=y$$

we have m $$a^{\frac{1}{n}}$$



and

$$\sqrt[n]{a}=y^{\frac{1}{m}}$$

and we get (III):

$$(\sqrt[n]{a})^m=y$$

and from (II) and (III):

$$(\sqrt[n]{a})^m=y=\sqrt[n]{a^m}$$

but I can't show that (IV):

$$(\sqrt[n]{a})^m=y=\sqrt[n]{a^m}=a^{\frac{m}{n}}$$

Since I have showed that

$$(\sqrt[n]{a})^m=\sqrt[n]{a^m}$$

I only need to show that

$$(\sqrt[n]{a})^m$$

is equal too

$$a^{\frac{m}{n}}$$

which is that

$$(a_1)^{\frac{1}{n}}(a_2)^{\frac{1}{n}}... (a_m)^{\frac{1}{n}}$$

(subscript is only for showing that it is m as)

is equal

$$a^{\frac{m}{n}}$$

in other words I need to show

$$(a)^{\frac{1}{n}}(a)^{\frac{1}{n}}=(a)^{\frac{2}{n}}$$

to show (IV) for any whole positive integer of n and m and any real number of a

then again how to prove (vv)

 

[micromass]

I guess this is my explenation for what I did get from the proof

http://bildr.no/view/1035536

But actually i wanted to prove that

$$\sqrt[n]{c^m}\sqrt[n]{d^m}=c^{m/n}d^{m/n}$$ (I)

dealing with integers when working with powers and roots are much simpler because it is easier to think about a number multiplied by itself n times and the nth roth of a number because it has to be a number one gets by doing that. If one could prove (I) then one could go from whole integers for n and m on the left side to a fraction on the right side and by that one would always see that one could first find a power and then a root even (or first root then power, to switch order of theese operations is provable) though it is written as the right side of (I)

The very definition is

$$c^{\frac{n}{m}}=\sqrt[m]{c^n}$$

So what you want is true by definition.

 

[I like Serena]

Isn't that:
$$(c^n)^{\frac{1}{m}}=\sqrt[m]{c^n}$$
?

 

[georg gill]

Isn't that:
$$(c^n)^{\frac{1}{m}}=\sqrt[m]{c^n}$$
?

If you say take the 22th power of 1.23. That is the same as multiplying 1.23 by itself 22 times. And if you then take the 6th rooth you take the rooth 6 times after each other and that always works for any positive real number. This is easy to understand but that this is the same as

$$1.23^{\frac{22}{6}}=1.23^{\frac{11}{3}}$$

is not as easy I think

 

[I like Serena]

And if you then take the 6th rooth you take the rooth 6 times after each other and that always works for any positive real number.

That is not the way it works.
To take the 6th root, you're looking for a number that if multiplied by itself 6 times will yield the original number.

Once you put up the power as a fraction (if that is allowed), the rules for fractions apply, and you can simplify the fraction.

 

[georg gill]

That is not the way it works.
To take the 6th root, you're looking for a number that if multiplied by itself 6 times will yield the original number.

Once you put up the power as a fraction (if that is allowed), the rules for fractions apply, and you can simplify the fraction.

Sorry I meant


To take the 6th root, you're looking for a number that if multiplied by itself 6 times will yield the original number.

I would really like to avoid calculation of powers and roots without whole integers. Is it possible to prove

$$(a)^{\frac{1}{n}}(a)^{\frac{1}{n}}=(a)^{\frac{2}{n}}$$

for any positvie integer by induction?