Timeline for Why Does Induction Prove Multiplication is Commutative?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Feb 20, 2013 at 0:13 | vote | accept | Russell Easterly | ||
| Feb 19, 2013 at 20:55 | answer | added | abo | timeline score: 6 | |
| Feb 19, 2013 at 4:07 | comment | added | Russell Easterly | Matrix addition and multiplication satisfy all of the axioms of Ring Theory (RT). Non-commutative rings are not models of RT+Ind where Ind is first order induction. Abo gives an example of a phi(x) we can prove using induction that is false in matrix arithmetic. | |
| Feb 18, 2013 at 22:23 | comment | added | Joel David Hamkins | Regarding your edit, Russell, isn't it clear that every countable set (whether finite or infinite) admits a successor function for which induction holds? Just pick a $0$, and then define $S(x)$ so that the $n^{th}$ element of the ring is $S^n(0)$. This will satisfy induction for the same reason that $\mathbb{N}$ satisfies induction. But of course, this $S$ will not interact with the ring operations meaningfully, and in the non-commutative case it cannot agree so as to make multiplication agree with the usual recursion, since then the inductive argument would make that operation commutative. | |
| Feb 18, 2013 at 22:04 | history | edited | Russell Easterly | CC BY-SA 3.0 |
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| Feb 18, 2013 at 2:40 | comment | added | Joel David Hamkins | If $S$ is meant to be a function symbol in first-order logic, then I agree with Joël that the displayed axiom follows from the axioms for equality. Following the link to Boucher's text, however, it seems that for GA he wants the axiom he calls PA3, which seems to assert that successor is functional (formalized as a binary relation giving the graph). | |
| Feb 18, 2013 at 1:43 | comment | added | Joël | Question from a bystander. I don't understand your successor axiom. Is it not a consequence of the general axioms of equality? | |
| Feb 17, 2013 at 15:57 | answer | added | abo | timeline score: 5 | |
| Feb 17, 2013 at 9:02 | comment | added | abo | I don't understand the "why" of your question. Consider 2 x 2 matrices. What do you propose to define as the successor relationship (from which you then define addition and multiplication)? What is the successor of 0? I imagine a first-order version cannot prove multiplication is commutative. | |
| Feb 17, 2013 at 4:23 | answer | added | Joel David Hamkins | timeline score: 6 | |
| Feb 17, 2013 at 3:48 | history | asked | Russell Easterly | CC BY-SA 3.0 |