4
$\begingroup$

Hi all I know that $5^1 = 5$, $5^2 = 25$, $5^3 = 125$.

But why is $5^{1.5} = 11.180339887498949$ ?

How did we get the number $11.180339887498949$ ?

$\endgroup$
11
  • 2
    $\begingroup$ $5^{1.5}=5^{1+0.5}=5\times5^{0.5}=5\sqrt{5}$. $\endgroup$ Jan 12, 2012 at 9:29
  • 2
    $\begingroup$ We know that, for example, $(5^2)^3 = 5^{2\times 3}$. So, whatever $5^{1.5}$ might be, we'd like it to satisfy $(5^{1.5})^2 = 5^{1.5\times 2} = 5^3 = 125$. So $5^{1.5}$ ought to be the square root of 125. $\endgroup$
    – user856
    Jan 12, 2012 at 9:30
  • 1
    $\begingroup$ To avoid any confusion about irrationals, consider instead $4^{1.5} = 8$. $\endgroup$ Jan 12, 2012 at 9:31
  • 1
    $\begingroup$ @Lieven, because I felt like the answer this question really ought to get is something long, comprehensive, and impressive, like André Nicolas' explanation of irrational exponents, and I didn't have time to write one. $\endgroup$
    – user856
    Jan 12, 2012 at 11:05
  • 3
    $\begingroup$ @Pacerier, how do we know that $\sqrt 5 = 2.23606...$? Well, 2 is too small because its square is 4, but 3 is too big because its square is 9, similarly 2.2 is too small, 2.3 is too big, 2.23 is too small, 2.24 is too big, and so on... Of course, there are more efficient methods. But this is something you could have invented yourself, and I find such explanations to be the most edifying. $\endgroup$
    – user856
    Jan 12, 2012 at 11:14

2 Answers 2

8
$\begingroup$

If $a,b$ are positive integers and $x > 0$, then $x^{a/b}$ is defined as $\sqrt[b]{x^a}$, where the $b^\text{th}$ root of $y>0$ is the unique positive real number $r$ such that $r^b = y$. So $5^{1.5}$ = $\sqrt{5^3}$, i.e. it's the number which squared is $125$.

This leaves open many questions such as "why do $n^\text{th}$ roots exist?" and "what about $x^\alpha$ where $\alpha$ is irrational?"

There are no easy answers to those questions which don't involve a first course in real analysis. If you haven't done university level real analysis, you kind of have to take on faith that exponentiation works and obeys the rules given. I know that's a disappointing answer, but it's the only decent answer I have.

$\endgroup$
3
  • $\begingroup$ But this is like a chicken-egg problem. So I understand that 5^1.5 = 5 × sqrt(5). But how do you solve for sqrt(5) in the first place? $\endgroup$
    – Pacerier
    Jan 12, 2012 at 11:00
  • $\begingroup$ Like I said: there is no easy answer. You either construct a sequence converging to $\sqrt{5}$ (think of succesively better approximate solutions to the equation $x^2 = 5$) or you realize it as the supremum of e.g. the set $\lbrace x \in \mathbb R \;|\; x^2 \leq 5\}$. Here's a note that explains it, but I think it's going to be way over the level you're at: PDF. It was not without reason I wrote that you have to take a lot of things on faith about exponentiation $\endgroup$
    – kahen
    Jan 12, 2012 at 11:10
  • $\begingroup$ Thanks for the link to that PDF Kahen! $\endgroup$ Jan 13, 2012 at 0:42
0
$\begingroup$

$$ 2\cdot2\cdot2=8, $$ so multiplying by $2$ three times is the same as multiplying by $8$,

and multiplying by $8$ one-third of one time is the same as multiplying by $2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .