Question

how can you prove that every positive integer can be expressed as the sum or difference of distinct powers of 3? We've got it down to a pattern we just dont know the summation notation for this type of problem

Answer 1

In similar way that you would prove every number has a unique binary extension. The summation notation would be something like\[\large \sum_{i=0}^n a_i \cdot 3^i\]where \(a_i\in \{0, 1, 2\}\)

Answer 2

*binary expansion

Answer 3

so a ∈ {-1,0,1}? because 3^0 = 1 ||| 3^1 - 3^0 = 2 ||| 3^1 = 3 and that 3^0 term always alternates between having a coefficient of 1, then -1, then 0...and so on

Answer 4

also the next power, 3^1, always alternates between having the coefficients -1,-1,-1,1,1,1,0,0,0....and 3^2 always alternates between having -1 as its coefficient 9 times, then 1 as the coefficient 9 times, and then 0.

Any more thoughts on how i could describe this?

Answer 5

Oh, I see. Yes. \(a\in\{-1, 0, 1\}\).

The summation notation would still be the same as what I had written. The actual summation notation would still be written the same way as I wrote it above.

Answer 6

However, if you could prove that every number is then expressible as a sum of distinct powers of three with \(a\in\{0, 1, 2\}\) you could make the observation that if you have a \(2\cdot3^n\) term, you can express this as \(3^{n+1}-3^n\). Hence, if you could prove that a unique trinary expansion exists for all positive integers, this would imply (with a bit more work) that every positive integer is expressible as the sum or difference of unique powers of three.