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4 ideas
15712 | 1 and 0, then add for naturals, subtract for negatives, divide for rationals, take roots for irrationals [Kaplan/Kaplan] |
Full Idea: You have 1 and 0, something and nothing. Adding gives us the naturals. Subtracting brings the negatives into light; dividing, the rationals; only with a new operation, taking of roots, do the irrationals show themselves. | |
From: R Kaplan / E Kaplan (The Art of the Infinite [2003], 1 'Mind') | |
A reaction: The suggestion is constructivist, I suppose - that it is only operations that produce numbers. They go on to show that complex numbers don't quite fit the pattern. |
15711 | The rationals are everywhere - the irrationals are everywhere else [Kaplan/Kaplan] |
Full Idea: The rationals are everywhere - the irrationals are everywhere else. | |
From: R Kaplan / E Kaplan (The Art of the Infinite [2003], 1 'Nameless') | |
A reaction: Nice. That is, the rationals may be dense (you can always find another one in any gap), but the irrationals are continuous (no gaps). |
15714 | 'Commutative' laws say order makes no difference; 'associative' laws say groupings make no difference [Kaplan/Kaplan] |
Full Idea: The 'commutative' laws say the order in which you add or multiply two numbers makes no difference; ...the 'associative' laws declare that regrouping couldn't change a sum or product (e.g. a+(b+c)=(a+b)+c ). | |
From: R Kaplan / E Kaplan (The Art of the Infinite [2003], 2 'Tablets') | |
A reaction: This seem utterly self-evident, but in more complex systems they can break down, so it is worth being conscious of them. |
15715 | 'Distributive' laws say if you add then multiply, or multiply then add, you get the same result [Kaplan/Kaplan] |
Full Idea: The 'distributive' law says you will get the same result if you first add two numbers, and then multiply them by a third, or first multiply each by the third and then add the results (i.e. a · (b+c) = a · b + a · c ). | |
From: R Kaplan / E Kaplan (The Art of the Infinite [2003], 2 'Tablets') | |
A reaction: Obviously this will depend on getting the brackets right, to ensure you are indeed doing the same operations both ways. |