0
$\begingroup$

Is it possible to have three line segments, $a$, $b$ and $c$, in an arrangement such that the geometry implicitly shows that the length of $c$ is equal to the length of $a$, raised to a power, which is the length of $b$? i.e. $c=a^b$

I'm looking for something similar to how you can show three segments, where $c=a \times b$ like this. I don't, however, want to use this to multiply $a$ by itself $b$ times. I have seen similar questions being answered this way, but it won't help in my case. I want the variable $b$ to be able to change without adding or removing geometry.

The geometry doesn't have to be constructable using just straightedge and compass.

$\endgroup$
2
  • $\begingroup$ @user90369 He already linked to that one saying it's not what he's looking for ("similar questions" in the question is the link). $\endgroup$
    – Arthur
    Commented Jul 20, 2017 at 11:05
  • $\begingroup$ @Arthur: Thanks. $\endgroup$
    – user90369
    Commented Jul 20, 2017 at 11:07

2 Answers 2

0
$\begingroup$

If you are not limited to straightedge and compass, you could then use the graph of a logarithmic function: in diagram below, you first draw points $U$, $A$ and $B$ such that $OU=1$, $OA=a$ and $OB=b$, then construct $C$ and $OC=c=a^b$. The construction should be clear from the diagram.

enter image description here

$\endgroup$
1
  • $\begingroup$ A solution using purely straight lines or even arc segments would have been preferable, but this definitely solves my problem. Much appreciated. $\endgroup$ Commented Jul 20, 2017 at 21:09
0
$\begingroup$

You can add two numbers geometrically along a line $\ell$ with chosen origin, and you can multiply two numbers along this line, when you have chosen a unit, i.e., fixed arbitrarily not only $0$, but also $1$ on $\ell$. (The ancient Greeks didn't come up with the latter idea, hence their products were always areas.)

But you cannot represent $a^b$ by a geometric "construction" (whether with ruler and compass or not), because $b$ cannot be viewed as a length, as in $c=a\cdot b$. In fact the native such $b$ is a positive integer that geometrically throws $a^b$ into $b$-dimensional space. Introducing a unit will then allow you to interpret $a^b$ as a length again. Various algebraic and analytic schemes are needed to interpret $a^b$ for $a>0$ and $b\in{\mathbb R}$ as a real number. But it is definitely impossible to arrive at $5^{\sqrt{2}}$ through a geometrical process.

$\endgroup$

You must log in to answer this question.

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