2 to the power of root 2

[apple2357]

Does 2 to the power of root 2 mean anything?
It doesn't make any sense to me when thinking about basic indices?

 

[Steven G]

Does 2 to the power of root 2 mean anything?
It doesn't make any sense to me when thinking about basic indices?

Yes 2^sqrt(2)does have meaning. Whey do you think otherwise?

 

[apple2357]

Because its irrational? And it was to the power of a rational number a/b , it makes sense to me , like 8^(2/3) = cube root and then square? But what about root 2? whats the equivalent interpretation?

 

[Dr.Peterson]

Because its irrational? And it was to the power of a rational number a/b , it makes sense to me , like 8^(2/3) = cube root and then square? But what about root 2? whats the equivalent interpretation?

You're right that rational powers can be explained in terms of the integer powers and roots that you initially learn about (x^3 = x*x*x; x^-3 = 1/x^3; x^(1/2) = square root of x; and so on).

When you come to irrational powers, you need to move into calculus: we define 2^sqrt(2) as the limit of 2^x as x approaches sqrt(2). That is, as you calculate 2^1, 2^1.4, 2^1.41, 2^1.414, and so on, you approach a specific number, which we take to be 2^sqrt(2). You just successively approximate the value.

You can also do this using logarithms.

 

[apple2357]

You're right that rational powers can be explained in terms of the integer powers and roots that you initially learn about (x^3 = x*x*x; x^-3 = 1/x^3; x^(1/2) = square root of x; and so on).

When you come to irrational powers, you need to move into calculus: we define 2^sqrt(2) as the limit of 2^x as x approaches sqrt(2). That is, as you calculate 2^1, 2^1.4, 2^1.41, 2^1.414, and so on, you approach a specific number, which we take to be 2^sqrt(2). You just successively approximate the value.

You can also do this using logarithms.

Thank you, that's very helpful. So we are now looking at things in a different way!

 

[Dr.Peterson]

Thank you, that's very helpful. So we are now looking at things in a different way!

Right. Once you get into irrational numbers, you can extend many concepts like exponents using an assumption of continuity, which makes them much more powerful.

 

[apple2357]

Right. Once you get into irrational numbers, you can extend many concepts like exponents using an assumption of continuity, which makes them much more powerful.

Thanks again. Is this the same thinking behind logs with negative bases or is that a simple No No?

 

[Dr.Peterson]

Thanks again. Is this the same thinking behind logs with negative bases or is that a simple No No?

Logs with negative bases are too messed up to be used at all. Consider that a negative base raised to an even power will be positive; to an odd power will be negative; to a fractional power like 1/2 will be non-real; to a fractional power like 1/3 will be negative; and so on. As a result, you can't even graph y = (-2)^x, so trying to find what x has to be to get a given value for y is just impossible. Nothing is continuous, or even continuously defined!

 

[Farzin]

You're right that rational powers can be explained in terms of the integer powers and roots that you initially learn about (x^3 = x*x*x; x^-3 = 1/x^3; x^(1/2) = square root of x; and so on).

When you come to irrational powers, you need to move into calculus: we define 2^sqrt(2) as the limit of 2^x as x approaches sqrt(2). That is, as you calculate 2^1, 2^1.4, 2^1.41, 2^1.414, and so on, you approach a specific number, which we take to be 2^sqrt(2). You just successively approximate the value.

You can also do this using logarithms.

Hi, how can we calculate 2 to the power of root 2 by using logarithms? Isn’t it the Gelfond - Schneider constant?

 

[Dr.Peterson]

Hi, how can we calculate 2 to the power of root 2 by using logarithms? Isn’t it the Gelfond - Schneider constant?

How does its having a name (and having been proved to be irrational) affect the question? We can't calculate it exactly, any more than we can calculate [MATH]\sqrt{2}[/MATH] exactly; but I don't think that was the question -- it was whether it is a number at all, as I read the OP. And clearly it is.

A calculator in effect uses logarithms to find a value for it.

 

[Farzin]

How does its having a name (and having been proved to be irrational) affect the question? We can't calculate it exactly, any more than we can calculate [MATH]\sqrt{2}[/MATH] exactly; but I don't think that was the question -- it was whether it is a number at all, as I read the OP. And clearly it is.

A calculator in effect uses logarithms to find a value for it.

Yes, I know that wasn’t the question and because I didn’t want to open a new thread, I just asked if it is possible to use logarithm to find approximate value of it. Thank you

 

[Dr.Peterson]

Here is one way: [MATH]2^\sqrt{2} = \left(e^{ln(2)}\right)^\sqrt{2} = e^{ln(2)\sqrt{2}}[/MATH]. Now evaluate that.

Or, of course, just use a calculator, which does at least something like that internally: [MATH]2^\sqrt{2} = 2.6651441426902251886502972498731...[/MATH]

 

[Farzin]

Here is one way: [MATH]2^\sqrt{2} = \left(e^{ln(2)}\right)^\sqrt{2} = e^{ln(2)\sqrt{2}}[/MATH]. Now evaluate that.

Or, of course, just use a calculator, which does at least something like that internally: [MATH]2^\sqrt{2} = 2.6651441426902251886502972498731...[/MATH]

Thank you.

 

[HallsofIvy]

Does 2 to the power of root 2 mean anything?
It doesn't make any sense to me when thinking about basic indices?

I've written this before but can't find it now so I will repeat.
I presume that you have been introduced to the power notation as $\displaystyle x^2= x\cdot x$, $\displaystyle x^3= x\cdot x\cdot x$, etc. with "$\displaystyle x^n$ mean "x multiplied by itself n times."

Obviously that requires that n be a positive integer. It just doesn't make sense to "multiply x by itself -1 times" or (your case) "multiply x by itself $\displaystyle \sqrt{2}$ times" or even "multiply x by itself 0 times".

But there are many times we would like to use them. And, hey, this is mathematics- everything depends on definitions. We are free to define things as we want. As long as everyone agrees we are free to define "$\displaystyle x^{\sqrt{2}}$" as we like.

Now, $\displaystyle x^n$, with n a positive integer, has two nice properties. One is that $\displaystyle (x^n)(x^m)= x^{n+m}$ because $\displaystyle x^n$ has n copies of x multiplied, $\displaystyle x^m$ has m copies, so $\displaystyle (x^n)(x^m)$ has n+ m copies of x multiplied. The other is that $\displaystyle (x^m)^n= x^{mn}$. To see that, imagine $\displaystyle x^m$ as a row of m x's. Taking that to the m power, we are multiplying m of those rows- think of a rectangle with m rows, each with n x's so there are mn x's.

As I said, we are free to define "$\displaystyle x^r$" as we please but it would be nice to do it so that those rules, $\displaystyle (x^n)(x^m)= x^{n+ m}$, and $\displaystyle (x^m)^n= x^{mn}$ were still true.

"0" is of course the "additive identity", n+ 0= n for any n. In particular $\displaystyle x^nx^0=x^{n+ 0}= x^n$ so we want $\displaystyle x^0= 1$ and that is how we define $\displaystyle x^0$ for any x.

​

Every positive integer has a "negative" such that $\displaystyle n+ (-n)= 0$ so we can write $\displaystyle x^{n}x^{-n}= x^{n- n}= x^0= 1$ so we want $\displaystyle x^{-n}$ to be the "multiplicative inverse" of $\displaystyle x^n$ which is $\displaystyle \frac{1}{x^n}$.​

​

That takes care of the integers. The next set of numbers is the rational numbers. A rational number is a number of the form $\displaystyle \frac{m}{n}$ where m and n are integers. Now we want to define x to a fractional power so that $\displaystyle (x^m)^n= x^{m+ n}$. In particular, we want $\displaystyle (x^{\frac{1}{n}})^n= x^{\frac{n}{n}}= x^1= x$. That means we want $\displaystyle x^{1/n} = \sqrt[n]{x}$, the nth root of x. Then, of course, $\displaystyle x^{\frac{m}{n}}= \left(x^{\frac{1}{n}}\right)^m= \left( \sqrt[n]{x} \right)^m= \sqrt[n]{x^m}$​

​

The irrational numbers, like your $\displaystyle \sqrt{2}$, cannot be defined "algebraically" as above- they have to be defined "analytically" with some kind of limiting process. The simplest is that every irrational number can be defined as the limit of a sequence of rational numbers. That is what we mean when we say that every irrational number iis For example, "$\displaystyle \pi$" is "3.141596..."- it is the limit of the sequence of rational numbers, 3, 3.1, 3.14, 3.141, 3.11415, etc. $\displaystyle \sqrt{2}= 1.4142...$. That is, $\displaystyle \sqrt{2}$ is the limit of the sequence of rational numbers 1, 1.4, 1.41, 1.414, 1.4142.​

​

And we define x to an irrational power to make $\displaystyle x^y$, for variable y, a continuous function. We define $\displaystyle x^{\sqrt{2}}$ as the limit of the sequence $\displaystyle x^1$, $\displaystyle x^{1.4}$, $\displaystyle x^{1.41}$, etc.​

​

Now, you can't actually calculate the limit of that sequence (and neither can your calculator) but you can approximate it by taking an approximation to $\displaystyle \sqrt{2}$. My calculator says that $\displaystyle \sqrt{2}$ is equal to 1.4142135623730950488016887242097. It isn't, of course, but that is the rational number approximation my calculator is using. And \(\displaystyle 2^\sqrt{2}\) is given as 2.6651441426902251886502972498731. Again, that is not the actual value of \(\displaystyle 2^\sqrt{2}\) but is the best approximation it can give.​

 

[joshuaa]

Yes 2^sqrt(2)does have meaning. Whey do you think otherwise?

Why

 

[Deleted member 4993]

Why

Does $\displaystyle \sqrt{2}$ make sense to you?

 

[Dr.Peterson]

Why

Note that that typo is over two years old.