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?
7=2^0 + 2^1 +2^2 (3 terms)
8=2^3 (1 term)
Will this affect the proof at all?
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.
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.