(a) (i) Copy and complete the addition \(\oplus\) and multiplication \(\otimes\) tables in modulo 5 on the set {2, 3, 4}. \(\oplus\) 2 3 4 2 3 4 \(\otimes\)...

(a)(i) Modulo 5 tables on {2, 3, 4}

Addition \(\oplus\) (add, then take the remainder on division by 5):

Multiplication \(\otimes\) (multiply, then take the remainder on division by 5):

(For example \(3\otimes 4 = 12 = 2\times 5 + 2 \equiv 2\), and \(4\oplus 3 = 7 = 5 + 2 \equiv 2\).)

(a)(ii)(a) Solve \(4\otimes e \oplus 2 \equiv 1 \ (\text{mod }5)\)

\(4\otimes e \equiv 1 \ominus 2 \equiv -1 \equiv 4 \ (\text{mod }5)\). So \(4e \equiv 4 \ (\text{mod }5)\). Multiplying both sides by the inverse of 4 (which is 4, since \(4\times 4 = 16 \equiv 1\)) gives \(e \equiv 16 \equiv 1 \ (\text{mod }5)\). Hence \(e = 1\).

(a)(ii)(b) Find n if \(4 \oplus n\otimes 2 \equiv 1 \ (\text{mod }5)\)

Multiplication first: \(n\otimes 2 = 2n\). Then \(4 \oplus 2n \equiv 1\), so \(2n \equiv 1 - 4 \equiv -3 \equiv 2 \ (\text{mod }5)\). Multiplying by the inverse of 2 (which is 3, since \(2\times 3 = 6 \equiv 1\)) gives \(n \equiv 6 \equiv 1 \ (\text{mod }5)\). Hence \(n = 1\).

(b) Validity of the statements (given \(p \Rightarrow q\))

Here p: Landi has cholera, q: Landi is in the hospital, and we are told \(p \Rightarrow q\) is true.

• (i) "If in hospital, then has cholera" is \(q \Rightarrow p\), the converse. It does not follow from \(p\Rightarrow q\), so it is not valid.
• (ii) "If not in hospital, then no cholera" is \(\lnot q \Rightarrow \lnot p\), the contrapositive of \(p\Rightarrow q\). It is logically equivalent, so it is valid.
• (iii) "If no cholera, then not in hospital" is \(\lnot p \Rightarrow \lnot q\), the inverse. It does not follow, so it is not valid.