Define addition of a set of symbols 01 by 000011101110 Defi

Define addition of a set of symbols {0,1} by : 0+0=0,0+1=1,1+0=1,1+1=0. Define multiplication by n where n is a non-negative integer by 0*0=0, 0*1=0, n*0=0, n*1=1+1+1+1+....+1(n terms) for n>0.

(a)Prove that if n is odd then n*1=1 and that if n is even then n*1=0.

(b)Verify that if one begins with the string a₀ a₁ the final digit is a₀+a_1.Verify that if one begins with the string a₀ a₁ a₂ the final digit is a₀+2a₁+a_2. Determine the final digit for each of the strings a₀ a₁ a₂ a₃, a₀ a₁ a₂ a₃ a₄, and a₀ a₁ a₂ a₃ a₄ a₅.

(c) Make a conjecture for a formular for the final digit of a₀ a₁ a₂....a_n. A proof is not required.

Thanks

2. The remaining questions concern the problem Sequence which was introduced in Home- work 6. Define addition of a set of symbols (0,1 by 1, 1 0 1, 1 1 0 0 0,0 1. Define multiplication by n where n is a non-negative integer by 1 for n 0 0, 0.1 0, n 0 0, m 1 n terms (a) Prove that if n is odd then n 1 1 and that if n is even then n 1 30 (b) Verify that if one begins with the string ao the final digit is ao ai. Verify that if one begins with the string ao ar a2 the final digit is ao 2a1 +a2. Determine the Bruni digit for each of the strings ao ni ma ma, on ai az aa atid as ar i asi e) Miske a conjectame for a foranula for final digit of un Aproof

Solution

Ans a)

For Odd , n.1 = 1+...1 (2k+1)times

2k+1=1+1+1....+1 (1 = (2k+1) times).

Since 1+1=0, then

2k+1=0+1+1....+1 ( 1 = (2k1) times).

But 0+1=1, so now

2k+1=1+...+1 ( 1 = (2k3) times)

   .
   .
   .
   .
   .
   .
   .
following this algorithm,

2k+1 = 0 + 1 = 1

=> n.1 =1 , when n is odd

........................................................................................................................................

For even , n.1 = 1+... +1 (2k)times

=> 2k=1+1+1 + ....+1 (we add 1 2k times).

Since 1+1=0, then

2k=0+1+1 +....+1 (we add 1 2k2 times).

=> 2k = 0 + 0 + 1 +....+1 ( 1 = (2k-4)times)
=> 2k = 0 + 0 + 0 + 1 +....+1 ( 1 = (2k-6times)

.
.
.
.
.
.
.
.
.
.
=> 2k = 0 + 0 + 0 +.....+0

But 0 + 0 = 0

=> 2k = 0 + 0 +.....+0
=> 2k = 0 +.....+0
.
.
.
.
.
.
.
=> 2k = 0

=> n.1 = 0 , when n is even