Now my only concern is this...
k/2 and (k-1)/2 here must be natural numbers (or 0), but k/2 and (k-1)/2 must very possibly NOT be the same number, then why can we use the same letter "m" in the upper limit of summation?
e.g.
7=2^0 + 2^1 +2^2 (3 terms)
8=2^3 (1 term)
Will this affect the proof at all?
I have some questions about this proof of uniqueness.
First we get that k1=j1.
Then repeating the same steps, we can show that k2=j2.
But the problem is how can we end this? j_r and k_l are not necessarily equal (r and l are not necessarily equal).
To prove uniqueness by contradiction, we assumed at the beginning that
To be completely general, r and l are not necessarily equal, so when we repeat your procedure of dividing by the leading term, at the end, something is going to be left over on one side of the equation but not the other. Say I assume r<l, then using your trick we can show that k1=j1, k2=j2, k_r=j_r, but we cannot say anything about j_(r+1), j_(r+2),...,j_l.
Ok, this is starting to get annoying. Suppose you have some number and we can represent as and . What I showed you was how to show that these two expressions must be equal. We first assume that WLOG since we could just reverse it otherwise. And thus we get [math1+2^{\jmath_2-\jmath_1}+\cdots=2^{k_1-\jmath_1}+\cdots[/tex], since by assumption we can see that except for the first term on each side the terms are going to be even. Thus, the left side is odd, and consequently the right side must be odd. But, how we set it up this is only going to be true if . So, we know that and thus . So, we may subtract them from both sides to get and we follow the exact same process until we have that all the exponents must be equal
WLOG, I assume r<l, then we can show that j1=k1, j2=k2, j3=k3, ..., j_r=k_r, and we can subtract 2^j1 from both sides, then subtract 2^j2 from both sides, then subtract 2^j3 from both sides, ..., then subtract 2^(j_r)from both sides. I understand everything up to here. What's giving me trouble is the justification AFTER this.
Then we will have some terms left over on the right hand side, namely 2^[k_(r+1)], 2^[k_(r+2)],..., 2^(k_l), and we can't use the same arguments.