Math Problem Statement

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).

Solution

To illustrate how floating-point arithmetic can violate associativity, we will choose a base-10 floating-point system with precision t=2t = 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)+ca+(b+c)(a + b) + c \neq a + (b + c) Choose:

  • a=1.2a = 1.2
  • b=1.5b = 1.5
  • c=0.01c = 0.01

Now, let's compute both sides.

  1. Left-hand side (a+b)+c(a + b) + c: a+b=1.2+1.5=2.7(no rounding needed)a + b = 1.2 + 1.5 = 2.7 \quad (\text{no rounding needed}) Then: (a+b)+c=2.7+0.01=2.71(rounded to two decimal places: 2.7)(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)a + (b + c): b+c=1.5+0.01=1.51(rounded to 1.5)b + c = 1.5 + 0.01 = 1.51 \quad (\text{rounded to } 1.5) Then: a+(b+c)=1.2+1.5=2.7a + (b + c) = 1.2 + 1.5 = 2.7

So, in this case: (a+b)+c=2.7anda+(b+c)=2.7(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 = 2.7
  • a+(b+c)=2.6a + (b + c) = 2.6 Now the associativity fails.

Part (b) - Multiplication:

We want to find an example where: (a×b)×ca×(b×c)(a \times b) \times c \neq a \times (b \times c) Choose:

  • a=1.2a = 1.2
  • b=3.5b = 3.5
  • c=0.1c = 0.1

Now, let's compute both sides.

  1. Left-hand side (a×b)×c(a \times b) \times c: a×b=1.2×3.5=4.2(no rounding needed)a \times b = 1.2 \times 3.5 = 4.2 \quad (\text{no rounding needed}) Then: (a×b)×c=4.2×0.1=0.42(rounded to two decimal places: 0.42)(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×(b×c)a \times (b \times c): b×c=3.5×0.1=0.35(no rounding needed)b \times c = 3.5 \times 0.1 = 0.35 \quad (\text{no rounding needed}) Then: a×(b×c)=1.2×0.35=0.42a \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.

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Math Problem Analysis

Mathematical Concepts

Floating Point Arithmetic
Precision
Associativity
Rounding Errors

Formulas

(a + b) + c ≠ a + (b + c)
(a * b) * c ≠ a * (b * c)

Theorems

Associative Property of Addition
Associative Property of Multiplication

Suitable Grade Level

Undergraduate (College-Level)