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Exercise 30 - Chapter 2

edited June 2018

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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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 \$$? 
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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.
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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 \$$.