Does it seem right to calculate fractional powers by walking steps method?

Question

We know that $x^y$ means $x$ multiplied by itself $y$ times or $x$ taking $y$ steps.

Is this method suitable when $y$ (the power) is fractional?

e.g.
if $x=2$ and $y = 3.5$, we would have
$$2^{3.5} = (2 \times 1) \times (2 \times 1) \times (2 \times 1) \times (2 \times 0.5) = 2 \times 2 \times 2 \times 1 = 8$$

The result is
$2^3 = 2^{3.5}$ (not possible)

I can't get the reason for it being the wrong method of calculating fractional powers.

Any help would be appreciated.

Answer 1

It doesn't even work for integer values. If it did we would be able to calculate $2^1$ by taking two "half-steps", i.e. $2^1\overset{?}{=}(2\times 0.5)\times(2\times 0.5)=1$.

This doesn't work because halving $x$ is only the right thing to do if you wanted to add half a time. It's not right for multiplying half a time. But we can make this idea work by taking logs, which reduces multiplying to adding. Now we get

$$\log (2^{3.5})=\log 2+\log 2+\log 2+0.5\log 2,$$ which is correct - and clearly different to $3\log 2=\log 8$.

So basically the issue is that (except for specific values of $x$), $0.5\log x$ is not the same as $\log (0.5x)$.

Answer 2

$2^n$ does not mean $\underbrace{(2\times 1)\times(2\times 1)\times\cdots\times(2\times1)}_{n\text{ times}}$ more than $2\times n$ means $\underbrace{(2+0)+\cdots+(2+0)}_{n\text{ times}}$: it is certainly true, but it's the kind of truth which may confuse talented cricket players.

For $n\in\Bbb N$, by definition $2^n=\underbrace{2\times\cdots\times2}_{n\text{ times}}$, and therefore, if anything, the relevant distributive identity is $2^{n+m}=2^n\times 2^m$, which would then become $2^n=\underbrace{2^1\times\cdots\times 2^1}_{n\text{ times}}$. Preservation, if possible, of this distributive property hints towards having $2^{-n}$ be the multiplicative inverse $2^n$ (so that $2^n\times 2^{-n}=2^{n+(-n)}=2^0=1$) and having $2^{0.5}$ be the (positive) real number such that $2^{0.5}\times 2^{0.5}=2$.