Solution.
Q1.
a)
i)
U={x:xβZ,1β€xβ€12}={1,2,3,4,5,6,7,8,9,10,11,12},A={2x:xβUandΒ xΒ isΒ aΒ multipleΒ ofΒ 4}={8,16,24},B={x:xβU,xdividesΒ 2}={2,4,6,8,10,12},C={x:xβU,x2β€16}={1,2,3,4}.
ii)
A={1,2,3,4,5,6,7,9,10,11,12},Aβ©B={2,4,6,10,12},Cβ(Aβ©B)={1,3}.
iii)
C={5,6,7,8,9,10,11,12},BβC={2,4,5,7,9,11}.
b)
1+5+9+β¦+(4nβ3)=n(2nβ1)forΒ allΒ nβ₯1.
Let
P(n):1+5+9+...+(4nβ3)=n(2nβ1),forΒ allΒ nβ₯1.P(1):1=1(2β
1β1),1=1,whichΒ isΒ true.P(1)isΒ true.AssumeΒ thatΒ P(n)Β isΒ trueΒ forΒ n=k.P(k):1+5+9+...+(4kβ3)=k(2kβ1)Β isΒ true.ProveΒ P(k+1)Β isΒ true.P(k+1):1+5+9+...+(4kβ3)+4(k+1)β3==k(2kβ1)+4(k+1)β3=2k2βk+4k+4β3==2k2+3k+1=2k2+2k+k+1=(k+1)(2k+1)==(k+1)(2k+1+1β1)=(k+1)(2(k+1)β1).So,Β P(k+1)Β isΒ true,Β wheneverΒ P(k)Β isΒ true,Β henceΒ P(n)isΒ true.
c)
x=866Β andΒ y=732.
i)
866=2β
433,732=2β
2β
3β
61=22β
3β
61,
from here the greatest common divisor ofΒ xΒ andΒ yΒ isΒ 2.
GCD(866,732)=2.
ax+by=β71β
866+84β
732=2.
ii)
866=2β
433,732=2β
2β
3β
61=22β
3β
61,
from here LCM(866,732)=2β
433β
2β
3β
61=361956.