Answer
See explanation
Work Step by Step
### Part (a): Uniqueness of Remainder (Proof by Contradiction)
**Statement:**
For any integer \( n \), it is impossible for
\[
n = 3q_1 + r_1 \quad \text{and} \quad n = 3q_2 + r_2,
\]
where \( q_1, q_2, r_1, r_2 \) are integers with
\[
0 \le r_1 < 3,\quad 0 \le r_2 < 3,\quad \text{and} \quad r_1 \neq r_2.
\]
**Proof:**
Assume, for the sake of contradiction, that there exists an integer \( n \) and integers \( q_1, q_2, r_1, r_2 \) satisfying
\[
n = 3q_1 + r_1 \quad \text{and} \quad n = 3q_2 + r_2,
\]
with \( 0 \le r_1, r_2 < 3 \) and \( r_1 \neq r_2 \).
Subtract the two representations of \( n \):
\[
3q_1 + r_1 = 3q_2 + r_2.
\]
Rearrange:
\[
3(q_1 - q_2) = r_2 - r_1.
\]
The left side is clearly divisible by 3. Therefore, the right side \( r_2 - r_1 \) must be divisible by 3.
Since \( r_1 \) and \( r_2 \) are each in the set \(\{0, 1, 2\}\), the possible differences \( r_2 - r_1 \) are \(-2\), \(-1\), \(0\), \(1\), or \(2\). The only multiple of 3 among these is 0. Hence, we must have:
\[
r_2 - r_1 = 0 \quad \Longrightarrow \quad r_2 = r_1.
\]
But this contradicts our assumption that \( r_1 \neq r_2 \). Thus, our original assumption is false, and it is impossible for \( n \) to have two distinct remainders upon division by 3.
---
### Part (b): If \(n^2\) Is Divisible by 3, Then \(n\) Is Divisible by 3
**Statement:**
For all integers \( n \), if \( n^2 \) is divisible by 3 then \( n \) is divisible by 3.
**Proof (by Contradiction and Division into Cases):**
Assume, for contradiction, that there exists an integer \( n \) such that \( n^2 \) is divisible by 3 but \( n \) is not divisible by 3.
By the quotient-remainder (division algorithm) theorem, every integer \( n \) can be written as
\[
n = 3q + r, \quad \text{where } r \in \{0, 1, 2\}.
\]
Since we assume \( n \) is not divisible by 3, the remainder \( r \) is either 1 or 2.
**Case 1:** \( n = 3q + 1 \)
Then,
\[
n^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1.
\]
Thus, \( n^2 \) leaves a remainder 1 when divided by 3 and is not divisible by 3.
**Case 2:** \( n = 3q + 2 \)
Then,
\[
n^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1.
\]
Again, \( n^2 \) leaves a remainder 1 when divided by 3 and is not divisible by 3.
In both cases, if \( n \) is not divisible by 3, then \( n^2 \) is not divisible by 3. This contradicts our assumption that \( n^2 \) is divisible by 3. Therefore, if \( n^2 \) is divisible by 3, \( n \) must be divisible by 3.
---
### Part (c): Prove That \(\sqrt{3}\) Is Irrational
**Proof by Contradiction (Using the Result of Part (b)):**
Assume, for the sake of contradiction, that \(\sqrt{3}\) is rational. Then we can write
\[
\sqrt{3} = \frac{p}{q},
\]
where \( p \) and \( q \) are integers with no common factors (i.e. the fraction is in lowest terms) and \( q \neq 0 \).
Squaring both sides gives:
\[
3 = \frac{p^2}{q^2} \quad \Longrightarrow \quad p^2 = 3q^2.
\]
Thus, \( p^2 \) is divisible by 3. By the result of part (b), if \( p^2 \) is divisible by 3 then \( p \) is divisible by 3. Therefore, there exists an integer \( k \) such that:
\[
p = 3k.
\]
Substitute \( p = 3k \) into the equation \( p^2 = 3q^2 \):
\[
(3k)^2 = 3q^2 \quad \Longrightarrow \quad 9k^2 = 3q^2 \quad \Longrightarrow \quad q^2 = 3k^2.
\]
Thus, \( q^2 \) is divisible by 3, and again by the result of part (b), \( q \) is divisible by 3.
Now both \( p \) and \( q \) are divisible by 3, contradicting the assumption that \(\frac{p}{q}\) is in lowest terms. This contradiction shows that our original assumption (that \(\sqrt{3}\) is rational) must be false.
Therefore, \(\sqrt{3}\) is irrational.
