(a)
(i) Copy and complete the addition \(\oplus\) and multiplication \(\otimes\) tables in modulo 5 on the set {2, 3, 4}.
(ii) Use the tables to:
(a) solve the equation \(4 \otimes e \oplus 2 \equiv 1 \pmod{5}\):
(b) find the value of n, if \(4 \oplus n \otimes 2 \equiv\) (mod 5).
(b) Consider the following statements:
p: Landi has cholera,
q: Landi is in the hospital.
If p = q, state whether or not the following statements are valid:
(i) If Landi is in the hospital, then he has cholera.
(ii) If Landi is not in the hospital, then he does not have cholera.
(iii) If Landi does not have cholera, then he is not in the hospital.
(a)(i) Modulo 5 tables on {2, 3, 4}
Addition \(\oplus\) (add, then take the remainder on division by 5):
| \(\oplus\) | 2 | 3 | 4 |
|---|
| 2 | 4 | 0 | 1 |
| 3 | 0 | 1 | 2 |
| 4 | 1 | 2 | 3 |
Multiplication \(\otimes\) (multiply, then take the remainder on division by 5):
| \(\otimes\) | 2 | 3 | 4 |
|---|
| 2 | 4 | 1 | 3 |
| 3 | 1 | 4 | 2 |
| 4 | 3 | 2 | 1 |
(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.
(a)(i) Modulo 5 tables on {2, 3, 4}
Addition \(\oplus\) (add, then take the remainder on division by 5):
| \(\oplus\) | 2 | 3 | 4 |
|---|
| 2 | 4 | 0 | 1 |
| 3 | 0 | 1 | 2 |
| 4 | 1 | 2 | 3 |
Multiplication \(\otimes\) (multiply, then take the remainder on division by 5):
| \(\otimes\) | 2 | 3 | 4 |
|---|
| 2 | 4 | 1 | 3 |
| 3 | 1 | 4 | 2 |
| 4 | 3 | 2 | 1 |
(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.
