How do you proof that for #a,b,cinRR#, #a/b=b/c=c/a<=>a=b=c#?
The reverse direction is trivial, assuming we add the condition that
For the forward direction, note first that all of
Now, there are two possible cases in which the claim would not hold. In the first case, one differs from the other two. Without loss of generality, suppose
In the second case, they all differ. Then, without loss of generality, suppose
Thus the only way to avoid a contradiction is if
Here's an alternative proof not using proof by contradiction...
Let me try to prove this without using contradiction:
To show in the other direction, suppose
So we find:
#a = bk = ck^2 = ak^3#
Dividing both ends by
#k^3 = 1#
Since we are told that
Notice my comment above about the Real cube root. If we were not told that