Lemma 3.1. For any right-derivable number pF(a1), pF(a100)=2pF(a1).

Proof: Since a e Q, a1 may be rewritten as ba1 for a e{0,1} , b e Q. That is, we may let pF(ba100)=pF(a100). We have:

pF(a100) =

pF(ba100) = pF(ba10) + pF(ba) [Defn. 1.7]

= [pF(ba1)+pF(b)] + pF(ba) [Defn. 1.7]

= pF(ba1)+[pF(b) + pF(ba)] [Assoc.]

= pF(ba1)+pF(ba1) [Defn. 1.7]

= 2pF(ba1)

Lemma 3.2. For any right derivable number pF(a1), pF(a1000)=3pF(a1).

Proof:

pF(a1000)= pF(a100) + pF(a1) [Defn. 1.7]

= 2pF(a1) + pF(a1) [Lemma 3.2]

= 3pF(a1).

Lemma 3.3. For any right derivable number pF(a1), pF(a10001)=5pF(a1).

Proof: (using lemmas 3.1 and 3.2)

pF(a10001)= pF(a1000) + pF(a100)

= 3pF(a1) + 2pF(a1).

back

next