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July 23

Vans

Why does Vans contain something that looks like a square root symbol? GeoffreyT2000 (talk) 02:04, 23 July 2015 (UTC)[reply]

From [1] and [2] it can be inferred that there is no special reason or relation to a square root - it's just an extension of the V.

PS. I haven't heard of Vans before, and it would have been easier for me to know what you're asking if you had put a link. -- Meni Rosenfeld (talk) 07:13, 23 July 2015 (UTC)[reply]

As anyone from the 1980s can affirm, it is because Vans are radical.OldTimeNESter (talk) 18:56, 24 July 2015 (UTC)[reply]

This is far from maths, but in the photo on this page, you can see a box from "Van Doren" with the same stylized V. As noted in our article, the company was founded by Paul Van Doren and others as The Van Doren Rubber Company, in 1966, so presumably the boxes in that photo are from that time frame. The full interview at that link is an interesting read, if that's your type of thing. --LarryMac | Talk 21:00, 24 July 2015 (UTC)[reply]

Note that the Van Halen logo is also like a backwards square root symbol (with extra parts added). Just coincidence, in both cases. StuRat (talk) 15:50, 25 July 2015 (UTC)[reply]

A recurring theme, apparently; I know it from the Dutch brand of Viking ice skates. —Oliphaunt (talk) 18:08, 25 July 2015 (UTC)[reply]

I wonder why that name is reversed, in the top pic. (I'd just think the picture had been reversed, but then the "Custom Made" tag would be backwards.) NOTE: They seem to alternate the pic at the top of the page, so you might have to click multiple times to find the one I mean. StuRat (talk) 19:23, 25 July 2015 (UTC)[reply]

Reminds me of the Tandon Corporation sign, in which the bar of the T was extended over the whole word, though interrupted by the riser of d; my then coworker said, "Oh, I thought the bar meant it's pronounced Tandōn." —Tamfang (talk) 19:36, 25 July 2015 (UTC)[reply]

Finitely (co)generated modules

A module M is finitely cogenerated if and only if Soc(M) is finitely generated and essential in M. Is the dual statement also true, i.e. M is finitely generated if and only if Rad(M) is superfluous in M and M/Rad(M) is finitely cogenerated? GeoffreyT2000 (talk) 20:30, 23 July 2015 (UTC)[reply]