Possible to explain non-integer indices?

thepikminman

I can't wrap my head around this, but can someone explain to me what the meaning of a number to the power of a non-integer number is? Obviously 10^4 means "10 multiplied by itself 4 times", but what would, say, 10^3.5 mean? Ten multiplied by itself 3.5 times? See I can't imagine that in my head. Maybe my initial explanation of indices is wrong. Any help?

Moody_mona

It might be worth considering it as 10^7/2.

When a number is raised to the power of a fraction like this, the numerator serves as a power and the denominator acts as a root. So this would be the square root of 10, all to the power of 7. (order doesn't matter so you could also consider it 10 to the power of 7, and then the square root of that.

Likewise 5^4/3 would be the cubed root of 5, all to the power of 4.

If you need a way to remember this, a nice idea is that the root is under the line of the fraction, just like tree roots are under the ground.

thepikminman

Moody_mona wrote: »

It might be worth considering it as 10^7/2.

When a number is raised to the power of a fraction like this, the numerator serves as a power and the denominator acts as a root. So this would be the square root of 10, all to the power of 7. (order doesn't matter so you could also consider it 10 to the power of 7, and then the square root of that.

Likewise 5^4/3 would be the cubed root of 5, all to the power of 4.

If you need a way to remember this, a nice idea is that the root is under the line of the fraction, just like tree roots are under the ground.

Thasks for the explanation but that kind of only explains how I would calculate it on paper. I'm looking for it to make sense intuitive if that's even possible. Like say it wasn't a fraction say it was 47^4.6345 what would the meaning of that be?

TheOneWhoDraws

There are a few ways of considering this, either you can consider it in terms of power series by considering e^x, and writing a real number r^p as e^(p*ln(r)). You an define ln(r) in a non-circular manner, and substitute this expression this into the power series of e^x to define r^p, where p is irrational.

Another, probably simpler way, is that you can get a sequence of rational numbers (since your question is pertaining more to irrational powers) which converge to the irrational number. You then raise some number r to this sequence, and as the sequence goes to infinity, r^(sequence) will converge to r^(irrational power). This is probably simpler, so here's an illustrative example.

What does it mean to have 2^pi?

Remember pi =3.14159...

Consider the sequence of numbers (3, 3.1, 3.14, 3.141, 3.1415, ...). This sequence, as the number of terms in it goes to infinity, is going to tend towards pi.

So you can now consider the sequence of numbers: (2^3, 2^3.1, 2^3.14, 2^3.141, 2^3.1415,...).

This sequence, as the amount of terms gets larger and larger (and the amount of terms in it goes towards infinity) is going to get closer and closer to 2^pi. We take the limit of this sequence as n goes to infinity (where n is the number of terms) as what we say when we mean 2^pi.

I hope you'll find this to be a simple yet satisfying explanation of the idea as to what's going on.

EDIT: IF you're doing a university degree, you'll typically find the latter of these (although you may see both) in a Real Analysis class (or Calculus 1 class), and you'll likely see the former (more likely than in Real Analysis) if you do a class in complex analysis. You may also see either in a class relating to the construction of the real numbers.

thepikminman

TheOneWhoDraws wrote: »

There are a few ways of considering this, either you can consider it in terms of power series by considering e^x, and writing a real number r^p as e^(p*ln(r)). You an define ln(r) in a non-circular manner, and substitute this expression this into the power series of e^x to define r^p, where p is irrational.

Another, probably simpler way, is that you can get a sequence of rational numbers (since your question is pertaining more to irrational powers) which converge to the irrational number. You then raise some number r to this sequence, and as the sequence goes to infinity, r^(sequence) will converge to r^(irrational power). This is probably simpler, so here's an illustrative example.

What does it mean to have 2^pi?

Remember pi =3.14159...

Consider the sequence of numbers (3, 3.1, 3.14, 3.141, 3.1415, ...). This sequence, as the number of terms in it goes to infinity, is going to tend towards pi.

So you can now consider the sequence of numbers: (2^3, 2^3.1, 2^3.14, 2^3.141, 2^3.1415,...).

This sequence, as the amount of terms gets larger and larger (and the amount of terms in it goes towards infinity) is going to get closer and closer to 2^pi. We take the limit of this sequence as n goes to infinity (where n is the number of terms) as what we say when we mean 2^pi.

I hope you'll find this to be a simple yet satisfying explanation of the idea as to what's going on.

EDIT: IF you're doing a university degree, you'll typically find the latter of these (although you may see both) in a Real Analysis class (or Calculus 1 class), and you'll likely see the former (more likely than in Real Analysis) if you do a class in complex analysis. You may also see either in a class relating to the construction of the real numbers.

Jesus....I have no f-ing idea what you are talking about lol. Are you actually serious with that explanation? Because it seems excessively complicated and you're saying it's simple...

thepikminman

TheOneWhoDraws wrote: »

EDIT: IF you're doing a university degree, you'll typically find the latter of these (although you may see both) in a Real Analysis class (or Calculus 1 class), and you'll likely see the former (more likely than in Real Analysis) if you do a class in complex analysis. You may also see either in a class relating to the construction of the real numbers.

Yeah I'm in 3rd year mech. engineering, I've done applied maths I and II, calculus 1 2 and 3, maths methods etc but I don't get how what you're saying answers my question in any way. Can you explain with words? Like "ten to the power of 5.4689 is 10 multiplied by itself...." obviously not exactly like that, but in that type of format. I'm not looking for a mathematical proof here. Thanks for the effort though.

TheOneWhoDraws

thepikminman wrote: »

Yeah I'm in 3rd year mech. engineering, I've done applied maths I and II, calculus 1 2 and 3, maths methods etc but I don't get how what you're saying answers my question in any way. Can you explain with words? Like "ten to the power of 5.4689 is 10 multiplied by itself...." obviously not exactly like that, but in that type of format. I'm not looking for a mathematical proof here. Thanks for the effort though.

It's written like that not because it's a mathematical proof (it's not), but because there's no other real way to simplify it further without losing what is actually happening. You can't define it as something like "10 multiplied by itself Pi times", because that doesn't make sense. It's 'simple' in that it's not a rigorous definition, it's the basic idea of what is happening.

Basically, let's say you want 10 to the power of pi.

First, you take the number 10^3. (ten to the power of three)
Second, you take the number 10^3.1 (ten to the power of 3.1 [Moody Mona explains what his means])
Next, you take the number 10^3.14 (again, Moody Mona explained what it means when it's a rational number)

You keep on taking numbers in that manner. Pi has infinitely many decimal places, so you would have to get an infinite list of numbers to 'reach it' (which you won't). The 'final' Number will be 10^pi, so you define the value of 10^pi to be the limit of the sequence (you should have done limits in Calculus but if not you'll need background reading in what they are and how to find this).

If you want 10 to the power of 5.4689. Observe that you can write this as a fraction, 54689 divided by 1000. So you take the '1000th root' of ten (what multiplied by ten one thousand times gives you 10?) and then you multiply that number by itself 54689 times, just like Moody Mona mentioned. It doesn't matter how many digits you take, if the number is rational (you can write it as a fraction of integers) you can do this. It's only when the number is irrational you have to take a long list of numbers and see what it would be f you could take infinitely many.

Victor

thepikminman wrote: »

Thasks for the explanation but that kind of only explains how I would calculate it on paper. I'm looking for it to make sense intuitive if that's even possible. Like say it wasn't a fraction say it was 47^4.6345 what would the meaning of that be?

OK, I presume you can get:
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
and so on.

So, imagine I'm digging column foundations for a square building.

Each foundation is 2 x 2 x 2 metres.

On a plan of the building, there are two rows and two columns of foundations.

So, the amount of concrete I need is 2^5 = 32 m3.

If I was using 3 x 3 x 3 metres foundations, I would need 3^3 x 2^2 = 108 m3. An so on.

---

Now, when doing fancy stuff, like fitting a line to some data on a graph, it doesn't always fit the form of n^p where p is a whole number, but we might want to still express things in relation to n.

On the attached graphs R^2 is a measure of how well the line fits the data. The maximum is 1 (perfect modelling) and the minimum is 0 (no relation whatsoever - data completely random). In image 1, the ratio is 0.3286, which is a poor score. However, in image 2, the ratio is 0.197, which is even worse. So, in this case, the best model is y = 10.805x^0.2237.

Victor

If I expand my data set and I again try the different models. I find that a polynomial of order 6, i.e. x +x^2 + x^3 ... x^6 fits my data really well - R^2 = 0.9735. This model uses integers.

However, that is only for this data set and an order of 6. Different data will mean a different 'best fit'.

prosaic

If I understand, you want to understand powers that are not integers, which includes [latex]x^\frac{1}{2}, x^\frac{2}{3}, x^\frac{3}{10}, x^{1.1}[/latex]

First, an exponent will have an integer part and a fraction (or decimal) part. Ignore irrational numbers for now (those like pi, that can't be expressed as a single fraction).
Example: [latex]3^{2.3}[/latex] is the same as [latex]3^2 \times 3^{0.3}[/latex]. Here using rule for splitting a power into pieces (e.g. [latex]3^3 = 3^{2+1} = 3^2 \times 3^1[/latex]).
The integer part is easy. [latex]3^2 = 9[/latex].
The decimal can be turned into a fraction 0.3 = 3/10. Again, use a rule for powers: [latex]3^\frac{3}{10} = 3^{\frac{1}{10} \times 3} = ({3^\frac{1}{10}})^3[/latex].
So, we just need to work out what [latex]3^\frac{1}{10}[/latex] is. That is the 10th root of 3. It's some number that multiplied by itself 10 times will give 3. The number is 1.116123 approx. Then [latex]1.116123 \times 1.116123 \times \dots (10 times) = 3[/latex].
[latex]3^{2.3} = 3 \times 3 \times 1.116123^3 = 9 \times 1.390389[/latex].

But how to get a visual grasp of it?
[latex]3^2[/latex] is the area of a square of side 3.
[latex]3^3[/latex] is the volume of a cube of side 3.

[latex]3^{2.5}[/latex] is what? It's not the volume of a cuboid of sides [latex]3 \times 3 \times \frac{3}{2}[/latex]. The third part is not a "full dimension" but a "half a dimension", whatever that means.

[latex]3^{0.5} = 3^\frac{1}{2}[/latex]. So, [latex]3 = 3^\frac{1}{2} \times 3^\frac{1}{2}[/latex]. It can be split into two pieces that multiply together to give the whole 3.
A cube of side 3 has volume [latex]3 \times 3 \times 3 = 3 \times 3 \times 3^\frac{1}{2} \times 3^\frac{1}{2} = 3^{2.5} \times 3^\frac{1}{2} = 3^{2.5} \times 1.73205[/latex].
The last 3 in [latex]3 \times 3 \times 3[/latex] has been split into two pieces: [latex]3 \times 3 \times (1.73205 \times 1.73205)[/latex]. Because [latex]3 \times 3 \times 3[/latex] is a multiplication, the last one is split, not into an addition of two pieces, but into a multiplication of two pieces.
So the "half a dimension" is done using a multiplication rather than an addition.

Instead of thinking of dimensions, it may be better to consider interest rates (compounded continuously) or bacteria growth rates as something you can visualise.
If an initial amount, x, will grow to 3 times it's size over time period t, after one period it will be [latex]x \times 3[/latex]. After time 2t, it will be [latex]x \times 3^2[/latex], etc. But after 2.5 periods, what will it be? It's mode of growth is multiplicative over time. Think of a colony of bacteria: at any time a portion of cells are splitting so over any time interval the growth will be a multiplication.
If the growth over any time interval of length [latex]\frac{t}{2}[/latex] is always the same; let's say r is the growth factor over [latex]\frac{t}{2}[/latex]. With amount x, after [latex]\frac{t}{2}[/latex], there will be [latex]x \times r[/latex] and after time t, there will be [latex]x \times r^2[/latex]. So [latex]r^2 = 3[/latex] and [latex]r = 3^\frac{1}{2}[/latex].
Then we can say that the initial x will grow over time 2.5t to [latex]x \times 3^{2.5}[/latex].
Likewise, after time 2.1t, the amount will be [latex]x \times 3^{2.1} = x \times (3^2 \times 3^{0.1})[/latex] where [latex]3^{0.1}[/latex] is the tenth root of 3.

Any decimal power can be split into pieces.
[latex]3^{2.134} = 3^2 * 3^{0.1} \times 3^{0.03} \times 3^{0.004} = 3^2 \times 3^{0.1} \times (3^{0.01})^3 \times (3^0.001)^4[/latex].
[latex]3^{0.01}[/latex] is the hundredth root of 3 and [latex]3^{0.001}[/latex] is the thousandth root of 3.

Rambling but I hope you get the point.

prosaic

I see that latex formatting doesn't work right on mobile version so I'll post again without the latex markup:

If I understand, you want to understand powers that are not integers, which includes x^(1/2), x^(2/3), x^(3/10), x^1.1

First, an exponent will have an integer part and a fraction (or decimal) part. Ignore irrational numbers for now (those like pi, that can't be expressed as a single fraction).
Example: 3^2.3 is the same as 3^2 * 3^0.3. Here using rule for splitting a power into pieces (e.g. 3^3 = 3^(2+1) = 3^2 * 3^1).
The integer part is easy. 3^2 = 9.
The decimal can be turned into a fraction 0.3 = 3/10. Again, use a rule for powers: 3^(3/10) = 3^((1/10)*3) = (3^(1/10))^3.
So, we just need to work out what 3^(1/10) is. That is the 10th root of 3. It's some number that multiplied by itself 10 times will give 3. The number is 1.116123 approx. Then 1.116123 * 1.116123 * ... (10 times) = 3.
3^2.3 = 3 * 3 * 1.116123^3 = 9 * 1.390389.

But how to get a visual grasp of it?
3^2 is the area of a square of side 3.
3^3 is the volume of a cube of side 3.

3^2.5 is what? It's not the volume of a cuboid of sides 3 * 3 * (3/2). The third part is not a "full dimension" but a "half a dimension", whatever that means.

3^0.5 = 3^(1/2). So, 3 = 3^(1/2) * 3^(1/2). It can be split into two pieces that multiply together to give the whole 3.
A cube of side 3 has volume 3 * 3 * 3 = 3 * 3 * 3^(1/2) * 3^(1/2) = 3^2.5 * 3^(1/2) = 3^2.5 * 1.73205.
The last 3 in 3 * 3 * 3 has been split into two pieces: 3 * 3 * (1.73205 * 1.73205). Because 3 * 3 * 3 is a multiplication, the last one is split, not into an addition of two pieces, but into a multiplication of two pieces.
So the "half a dimension" is done using a multiplication rather than an addition.

Instead of thinking of dimensions, it may be better to consider interest rates (compounded continuously) or bacteria growth rates as something you can visualise.
If an initial amount, x, will grow to 3 times it's size over time period t, after one period it will be x * 3. After time 2t, it will be x * 3^2, etc. But after 2.5 periods, what will it be? It's mode of growth is multiplicative over time. Think of a colony of bacteria: at any time a portion of cells are splitting so over any time interval the growth will be a multiplication.
If the growth over any time interval of length (t/2) is always the same; let's say r is the growth factor over (t/2). With amount x, after (t/2), there will be x * r and after time t, there will be x * r^2. So r^2 = 3 and r = 3^(1/2).
Then we can say that the initial x will grow over time 2.5t to x * 3^2.5.
Likewise, after time 2.1t, the amount will be x * 3^2.1 = x * (3^2 * 3^0.1) where 3^0.1 is the tenth root of 3.

Any decimal power can be split into pieces.
3^2.134 = 3^2 * 3^0.1 * 3^0.03 * 3^0.004 = 3^2 * 3^0.1 * (3^0.01)^3 * (3^0.001)^4.
3^0.01 is the hundredth root of 3 and 3^0.001 is the thousandth root of 3.

Rambling but I hope you get the point.

prosaic

Another example of fractional powers occurs in musical scales.

Two notes, f2 and f1, are an octave apart when note f2 is twice the frequency of f1. f2 = 2*f1

In the standard musical scale, the octave is split into 12 semitones. The spacing in frequency of two notes that are one semitone apart is always the same. If f2 and f1 are one semitone apart then f2 = s*f1 where s is the constant factor increase in frequency.

Since there are 12 semitones in an octave, as you step up through one octave, you step up 12 times each time multiplying the frequency by s.
If f2 and f1 are an octave apart, f2 = 2*f1 = f1*s*s*s*...(12 times) = f1 * s^12. Therefore, s^12 = 2 or s = 2^(1/12). s is the twelfth root of 2 which is 1.059463.
Middle C is 261.63 Hz. C# = C*1.059463 = 277.183 Hz. D = C*(1.059463)^2 = 293.665. E = C*(1.059463)^4 = 329.628 etc.
E = C*2^(4/12) = C*2^(1/3) = C*2^(0.3333)

Bit cynical

I will have a go at an explanation. It is a bit long-winded but I think it explains it properly.

We are happy with the idea that

[latex]\displaystyle{a\times a\times a\times a\times a=a^{5}}[/latex].

We can also see that

[latex]\displaystyle{\left(a\times a\times a\right)\times\left(a\times a\right)=a^{3}\times a^{2}}[/latex],

and that the above is also [latex]\displaystyle{a}[/latex] multiplied by itself five times and therefore the same as [latex]\displaystyle{a^5}[/latex].

So [latex]\displaystyle{a^3\times a^2=a^5}[/latex].

The general rule is [latex]\displaystyle{a^b+a^c=a^{b+c}}[/latex].

So far so good. But what about

[latex]\displaystyle{a^{0.5}\times a^{0.5}}[/latex]? What is that? Well, applying the rule we find that

[latex]\displaystyle{a^{0.5}\times a^{0.5}=a^{0.5+0.5}=a^1}[/latex] which is just equal to [latex]\displaystyle{a}[/latex].

So [latex]\displaystyle{a^{0.5}}[/latex] is a number which, when multiplied by itself, gives [latex]\displaystyle{a}[/latex]. But this is the definition of the square root [latex]\displaystyle{\sqrt{a}\times\sqrt{a}=a }[/latex].

[latex]\displaystyle{a^{0.5}[/latex] or more commonly [latex]\displaystyle{a^{1/2}[/latex] is just another way of writing the square root of [latex]\displaystyle{a}[/latex].

By extension [latex]\displaystyle{a^{1/3}[/latex] is the cube root since

[latex]\displaystyle{a^{1/3}\times a^{1/3}\times a^{1/3}=a^{\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)}=a^{1}}[/latex].

So getting back to your original question: what does [latex]\displaystyle{10^{3.5}}[/latex] mean? Well [latex]\displaystyle{3.5=3+\frac{1}{2}}[/latex] and so applying the rule

[latex]\displaystyle{10^{3.5}=10^{3+\frac{1}{2}}=10^3\times 10^{1/2}=10^3+\sqrt{10}[/latex]. That is what it means.

But it is usually more convenient to leave it as [latex]\displaystyle{10^{7/2}}[/latex] as it is handier in further calculations.

Bit cynical

What is the meaning of ten to the power of 5.4689.

Well 5.4689 is 5 + 4589/10,000.

Lets take this in parts.

[latex]\displaystyle{10^{1/10,000}[/latex] is the ten thousandth root of 10.

[latex]\displaystyle{10^{4589/10,000}[/latex] is the (ten thousandth root of ten) to the power of 4589.

[latex]\displaystyle{10^{5+4589/10,000}[/latex] is ((ten thousandth root of ten) to the power of 4589) times (ten to the power of five).

So in purely verbal terms.

a. Calculate the ten thousandth root of ten
b. Raise that number to the power of 4589 and
c. multiply it by ten to the power of five

Not the best way of doing it as a lot of precision will be lost but in mathematical terms that is what it is.

thepikminman

Bit cynical wrote: »

I will have a go at an explanation. It is a bit long-winded but I think it explains it properly.

We are happy with the idea that

[latex]\displaystyle{a\times a\times a\times a\times a=a^{5}}[/latex].

We can also see that

[latex]\displaystyle{\left(a\times a\times a\right)\times\left(a\times a\right)=a^{3}\times a^{2}}[/latex],

and that the above is also [latex]\displaystyle{a}[/latex] multiplied by itself five times and therefore the same as [latex]\displaystyle{a^5}[/latex].

So [latex]\displaystyle{a^3\times a^2=a^5}[/latex].

The general rule is [latex]\displaystyle{a^b+a^c=a^{b+c}}[/latex].

So far so good. But what about

[latex]\displaystyle{a^{0.5}\times a^{0.5}}[/latex]? What is that? Well, applying the rule we find that

[latex]\displaystyle{a^{0.5}\times a^{0.5}=a^{0.5+0.5}=a^1}[/latex] which is just equal to [latex]\displaystyle{a}[/latex].

So [latex]\displaystyle{a^{0.5}}[/latex] is a number which, when multiplied by itself, gives [latex]\displaystyle{a}[/latex]. But this is the definition of the square root [latex]\displaystyle{\sqrt{a}\times\sqrt{a}=a }[/latex].

[latex]\displaystyle{a^{0.5}[/latex] or more commonly [latex]\displaystyle{a^{1/2}[/latex] is just another way of writing the square root of [latex]\displaystyle{a}[/latex].

By extension [latex]\displaystyle{a^{1/3}[/latex] is the cube root since

[latex]\displaystyle{a^{1/3}\times a^{1/3}\times a^{1/3}=a^{\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)}=a^{1}}[/latex].

So getting back to your original question: what does [latex]\displaystyle{10^{3.5}}[/latex] mean? Well [latex]\displaystyle{3.5=3+\frac{1}{2}}[/latex] and so applying the rule

[latex]\displaystyle{10^{3.5}=10^{3+\frac{1}{2}}=10^3\times 10^{1/2}=10^3+\sqrt{10}[/latex]. That is what it means.

But it is usually more convenient to leave it as [latex]\displaystyle{10^{7/2}}[/latex] as it is handier in further calculations.

Thanks for the attempt, but like i said, I know exactly how to calculate the value of and number raised to a non-integer. I know about adding the roots when multiplied, subtracting when dividing, splitting it up etc. I'm not asking how to do the problem, I'm just asking how to visualize it, if that's even possible.

thepikminman

Bit cynical wrote: »

What is the meaning of ten to the power of 5.4689.

Well 5.4689 is 5 + 4589/10,000.

Lets take this in parts.

[latex]\displaystyle{10^{1/10,000}[/latex] is the ten thousandth root of 10.

[latex]\displaystyle{10^{4589/10,000}[/latex] is the (ten thousandth root of ten) to the power of 4589.

[latex]\displaystyle{10^{5+4589/10,000}[/latex] is ((ten thousandth root of ten) to the power of 4589) times (ten to the power of five).

So in purely verbal terms.

a. Calculate the ten thousandth root of ten
b. Raise that number to the power of 4589 and
c. multiply it by ten to the power of five

Not the best way of doing it as a lot of precision will be lost but in mathematical terms that is what it is.

Thanks for the attempt, but like i said, I know exactly how to calculate the value of and number raised to a non-integer. I know about adding the roots when multiplied, subtracting when dividing, splitting it up etc. I'm not asking how to do the problem so I get the correct answer, I'm just asking how to visualize it, if that's even possible.

thepikminman

prosaic wrote: »

If I understand, you want to understand powers that are not integers, which includes [latex]x^\frac{1}{2}, x^\frac{2}{3}, x^\frac{3}{10}, x^{1.1}[/latex]

First, an exponent will have an integer part and a fraction (or decimal) part. Ignore irrational numbers for now (those like pi, that can't be expressed as a single fraction).
Example: [latex]3^{2.3}[/latex] is the same as [latex]3^2 \times 3^{0.3}[/latex]. Here using rule for splitting a power into pieces (e.g. [latex]3^3 = 3^{2+1} = 3^2 \times 3^1[/latex]).
The integer part is easy. [latex]3^2 = 9[/latex].
The decimal can be turned into a fraction 0.3 = 3/10. Again, use a rule for powers: [latex]3^\frac{3}{10} = 3^{\frac{1}{10} \times 3} = ({3^\frac{1}{10}})^3[/latex].
So, we just need to work out what [latex]3^\frac{1}{10}[/latex] is. That is the 10th root of 3. It's some number that multiplied by itself 10 times will give 3. The number is 1.116123 approx. Then [latex]1.116123 \times 1.116123 \times \dots (10 times) = 3[/latex].
[latex]3^{2.3} = 3 \times 3 \times 1.116123^3 = 9 \times 1.390389[/latex].

But how to get a visual grasp of it?
[latex]3^2[/latex] is the area of a square of side 3.
[latex]3^3[/latex] is the volume of a cube of side 3.

[latex]3^{2.5}[/latex] is what? It's not the volume of a cuboid of sides [latex]3 \times 3 \times \frac{3}{2}[/latex]. The third part is not a "full dimension" but a "half a dimension", whatever that means.

[latex]3^{0.5} = 3^\frac{1}{2}[/latex]. So, [latex]3 = 3^\frac{1}{2} \times 3^\frac{1}{2}[/latex]. It can be split into two pieces that multiply together to give the whole 3.
A cube of side 3 has volume [latex]3 \times 3 \times 3 = 3 \times 3 \times 3^\frac{1}{2} \times 3^\frac{1}{2} = 3^{2.5} \times 3^\frac{1}{2} = 3^{2.5} \times 1.73205[/latex].
The last 3 in [latex]3 \times 3 \times 3[/latex] has been split into two pieces: [latex]3 \times 3 \times (1.73205 \times 1.73205)[/latex]. Because [latex]3 \times 3 \times 3[/latex] is a multiplication, the last one is split, not into an addition of two pieces, but into a multiplication of two pieces.
So the "half a dimension" is done using a multiplication rather than an addition.

Instead of thinking of dimensions, it may be better to consider interest rates (compounded continuously) or bacteria growth rates as something you can visualise.
If an initial amount, x, will grow to 3 times it's size over time period t, after one period it will be [latex]x \times 3[/latex]. After time 2t, it will be [latex]x \times 3^2[/latex], etc. But after 2.5 periods, what will it be? It's mode of growth is multiplicative over time. Think of a colony of bacteria: at any time a portion of cells are splitting so over any time interval the growth will be a multiplication.
If the growth over any time interval of length [latex]\frac{t}{2}[/latex] is always the same; let's say r is the growth factor over [latex]\frac{t}{2}[/latex]. With amount x, after [latex]\frac{t}{2}[/latex], there will be [latex]x \times r[/latex] and after time t, there will be [latex]x \times r^2[/latex]. So [latex]r^2 = 3[/latex] and [latex]r = 3^\frac{1}{2}[/latex].
Then we can say that the initial x will grow over time 2.5t to [latex]x \times 3^{2.5}[/latex].
Likewise, after time 2.1t, the amount will be [latex]x \times 3^{2.1} = x \times (3^2 \times 3^{0.1})[/latex] where [latex]3^{0.1}[/latex] is the tenth root of 3.

Any decimal power can be split into pieces.
[latex]3^{2.134} = 3^2 * 3^{0.1} \times 3^{0.03} \times 3^{0.004} = 3^2 \times 3^{0.1} \times (3^{0.01})^3 \times (3^0.001)^4[/latex].
[latex]3^{0.01}[/latex] is the hundredth root of 3 and [latex]3^{0.001}[/latex] is the thousandth root of 3.

Rambling but I hope you get the point.

Thank you, you're the only one who actually understood my question. Everyone's so keen on just finding out the way to get to the final answer, they can't even comprehend the idea of trying to understand what's going on. (not being mean, it's just annoying though)

Anyway, I still don't really get it, but I appreciate the help. I understand most of what you're saying, but I'm starting to think it's like the 4th or higher dimension - impossible to visualize. But all the replies have proven to me that my question was actually a reasonable one, so that's good enough for me!

Michael Collins

There's always the (admittedly obvious) graphical method.

We can visualise the graph of the function

[latex] \displaystyle f(x) = 3^x [/latex]

as the following graph



The dots indicate the integer powers:

[latex] \displaystyle 3^{-2},,,,3^{-1},,,,3^{0},,,,3^{1}, ,,,3^{2} [/latex]

You could look at this, as prosaic says, as a growth rate - since you know how to calculate 3^x for non-integer powers, you can fill in what the function does between the integer powers.

Note that other posts have only explained things in terms of rationals, but there is a whole world out there of irrationals, that cannot be related to a number raised to an integer power, inside an integer-root, i.e. it cannot be written as

[latex] \displaystyle y = {^m}\sqrt{x^n} = x^{\frac{n}{m}} [/latex]

One such example would be

[latex] 3^{\pi} [/latex].

The graph above (if continued) would give a result for every possible real power, include irrational ones.

prosaic

thepikminman wrote: »

Anyway, I still don't really get it, but I appreciate the help. I understand most of what you're saying, but I'm starting to think it's like the 4th or higher dimension - impossible to visualize. But all the replies have proven to me that my question was actually a reasonable one, so that's good enough for me!

It is a bit of a head wrecker trying to think what it could be visualised as. Best I can come up with is multiplicative growth rates as in interest, bacteria and sound perception.

You could also think about decibel and richter scales. And logarithmic graph plotting in general. The human ear has an astounding range of sensitivity from minute sounds to high energy sound of a jet engine. The low threshold of hearing is 0 db and point of ear damaging sound level is around 120 db. There is an energy difference between these of 10^12. Every 10 db represents an increase in energy of 10 times. (I use the word energy instead of power as it might be easier to visualise (power = energy per second) )

A logarithmic scale is one in which the scale is zooming out as you go to higher values. If you have a log scale from -2 to 3, that covers values from 10^-2 = 0.01 to 10^3 = 1000. At the low end of the scale, very small changes of value can be seen. At the high end of the scale, very large changes of value are shown. For every step up of 1 unit, the scale zooms out by a factor of 10. If you step up 0.5 units on the scale, it zooms out by 10^0.5 = 3.162. If you step up by 0.1 on the scale, you zoom out by 10^0.1 = 1.2589.

Here, zoom is a multiplicative scaling factor. Zoom out by a factor of 10 can be broken up into 10 steps of smaller zoom-out. Each small zoom-out is a scaling of 10^0.1 or 1.2589. Do 10 of those small zoom-outs: 1.2589*1.2589*... (10 times) ... *1.2589 = 10. On the log scale these small steps are a 10th of a unit.

A log scale is handy for representing anything that ranges from very small values to very large values. It's also handy for showing exponential growth rates. The graph above, if drawn on a log scale would be a straight line.

chakotha

I think the denominator in an index that is a fraction is more a notational convenience really.

Multiplying a number by itself 1/2 times isn't going to give you the square root is it. It's meaningless.

But thinking of it as the root then things like 12^3/2 become readable and intuitive.