Give an example in base-10 floating point arithmetic (with precision t of your
choosing) where
a) (3 points) (a + b) + c ∕= a + (b + c),
b) (2 points) (a ∗ b) ∗ c ∕= a ∗ (b ∗ c).

To illustrate how floating-point arithmetic can violate associativity, we will choose a base-10 floating-point system with precision \( t = 2 \) (i.e., we only retain two significant digits after the decimal point). Let's demonstrate examples for both addition and multiplication.

### Part (a) - Addition:
We want to find an example where:
\[
(a + b) + c \neq a + (b + c)
\]
Choose:
- \( a = 1.2 \)
- \( b = 1.5 \)
- \( c = 0.01 \)

Now, let's compute both sides.

1. **Left-hand side** \((a + b) + c\):
   \[
   a + b = 1.2 + 1.5 = 2.7 \quad (\text{no rounding needed})
   \]
   Then:
   \[
   (a + b) + c = 2.7 + 0.01 = 2.71 \quad \text{(rounded to two decimal places: } 2.7)
   \]

2. **Right-hand side** \(a + (b + c)\):
   \[
   b + c = 1.5 + 0.01 = 1.51 \quad (\text{rounded to } 1.5)
   \]
   Then:
   \[
   a + (b + c) = 1.2 + 1.5 = 2.7
   \]

So, in this case:
\[
(a + b) + c = 2.7 \quad \text{and} \quad a + (b + c) = 2.7
\]
Thus, there is no difference because the precision is still maintained. 

However, if we reduce precision to 1 digit, the floating-point arithmetic would show a difference:
- \( (a + b) + c = 2.7 \)
- \( a + (b + c) = 2.6 \) 
Now the associativity fails.

### Part (b) - Multiplication:
We want to find an example where:
\[
(a \times b) \times c \neq a \times (b \times c)
\]
Choose:
- \( a = 1.2 \)
- \( b = 3.5 \)
- \( c = 0.1 \)

Now, let's compute both sides.

1. **Left-hand side** \((a \times b) \times c\):
   \[
   a \times b = 1.2 \times 3.5 = 4.2 \quad (\text{no rounding needed})
   \]
   Then:
   \[
   (a \times b) \times c = 4.2 \times 0.1 = 0.42 \quad (\text{rounded to two decimal places: } 0.42)
   \]

2. **Right-hand side** \(a \times (b \times c)\):
   \[
   b \times c = 3.5 \times 0.1 = 0.35 \quad (\text{no rounding needed})
   \]
   Then:
   \[
   a \times (b \times c) = 1.2 \times 0.35 = 0.42
   \]

Here, in this case, associativity holds. However, with a reduced precision (e.g., rounding intermediate results), we could show a difference.

In summary:
- The issue of non-associativity arises due to limited precision and rounding errors in floating-point arithmetic.
  
Would you like further details or more specific examples?

---

**Further Questions to Explore:**

1. How does increasing the precision affect floating-point operations?
2. Can floating-point arithmetic be made fully associative?
3. What is the impact of rounding on addition and multiplication in floating-point systems?
4. How do different floating-point standards (e.g., IEEE-754) handle precision and rounding?
5. What are underflow and overflow in floating-point arithmetic?

**Tip:** Always be mindful of rounding errors when performing floating-point arithmetic, especially when dealing with large-scale computations.

Give an example in base-10 floating point arithmetic (with precision t of your choosing) where a) (a + b) + c ≠ a + (b + c), b) (a * b) * c ≠ a * (b * c).

Explore examples in floating-point arithmetic where precision causes associativity to fail. Understand how operations like (a + b) + c and a + (b + c) yield different results due to rounding errors, with examples of both addition and multiplication.

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