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Exercise 30 - Chapter 2

Again taking the divisibility order \( ( \mathbb{N} , | ) \). Someone proposes 0 as the monoidal unit and + as the monoidal product.

Does that proposal satisfy the conditions of Definition 2.2?

Why or why not?

divisibility order: We write \( m|n \) to mean that m divides into n without remainder.

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Comments

  • 1.

    Consider \( x_1 = 2, x_2 = 3, y_1 = 4, y_2 = 9 \). Clearly \( x_1 | y_1 \) and \( x_2 | y_2 \) .

    Does \( 2 + 3 | 4 + 9 \)?

    Comment Source:Consider \\( x_1 = 2, x_2 = 3, y_1 = 4, y_2 = 9 \\). Clearly \\( x_1 | y_1 \\) and \\( x_2 | y_2 \\) . Does \\( 2 + 3 | 4 + 9 \\)?
  • 2.

    Fredrick: That's the same counterexample I settled on. Amusingly, my "pick a random set of numbers" approach kept finding ones that worked! (2|4, 3|6 => 5 | 10, for example).

    This is one demonstration of why just finding a few small examples to satisfy ourselves an assertion is true isn't sufficient. It's easy to mislead yourself into thinking you've stumbled upon a relation that doesn't actually exist.

    Comment Source:Fredrick: That's the same counterexample I settled on. Amusingly, my "pick a random set of numbers" approach kept finding ones that worked! (2|4, 3|6 => 5 | 10, for example). This is one demonstration of why just finding a few small examples to satisfy ourselves an assertion is true isn't sufficient. It's easy to mislead yourself into thinking you've stumbled upon a relation that doesn't actually exist.
  • 3.
    edited May 2018

    @JaredSummers My intuition told me to pick right hand values that are relatively prime. Does that have any relevance or did I just get lucky. In your example 4 and 6 share a common factor, 2, and so that gets rejected. The pairs \( 2|4 , 5|15 \rightarrow 2 + 5 | 4 + 15 = 7 | 19 \) seems to support the intuition as does \( 2|4 , 3|3 \rightarrow 2+3 | 4+3 = 5|7 \).

    Comment Source:@JaredSummers My intuition told me to pick right hand values that are relatively prime. Does that have any relevance or did I just get lucky. In your example 4 and 6 share a common factor, 2, and so that gets rejected. The pairs \\( 2|4 , 5|15 \rightarrow 2 + 5 | 4 + 15 = 7 | 19 \\) seems to support the intuition as does \\( 2|4 , 3|3 \rightarrow 2+3 | 4+3 = 5|7 \\).
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