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Saying "It takes surprisingly little time to do!" isn't trying to give a lower bound on how much time it takes. Obviously not completely unambiguous, but the "s
by tgb 6y ago
Saying "It takes surprisingly little time to do!" isn't trying to give a lower bound on how much time it takes. Obviously not completely unambiguous, but the "sufficient" reading is completely natural English.
- lostcolony 6y agoIt's not setting a bound at all. It -is- saying that the amount of time is "surprisingly little". But to the topic at hand, the logical implication is that "to do it" => "takes surprisingly little time" (that is, doing it implies you took surprisingly little time) Not "taking surprisingly little time" => "to do it" (that is, taking surprisingly little time implies you do it; after all, you might have taken surprisingly little time to do -something else-). Same as the above; the implication is that "success" implies "(at least) a surprisingly small number of adults were involved", not that "(at least) a surprisingly small number of adults were involved" implies "success". I find it helpful to use formal logic to explain stuff like this. If I make a statement that p => q (that is, p implies q), and that statement is true, it does NOT necessarily follow that q => p (that is, q implies p) is true (it may or may not be; it is independent from a formal logic perspective). It is necessary to evaluate q => p separately, in which case we may find it to be true, OR we may find q =/> p to be true (that is, q does not imply p; or put another way, the fact q is true tells us nothing about whether p is). As an example, "My shoes are wet because it is raining", p = it is raining, and q = my shoes are wet. p => q means if it is raining (p), my shoes are wet (q). But it does not also follow from that that q => p; that is, just because my shoes are wet (q) it does not mean it must be raining (p; I might have, after all, been out watering the lawn). Thus, q =/> p. For the tango example, p = a tango is possible, q = there are two people. p => q, but q =/> q (that is, two people does not imply a tango is possible). In this case, p = foster child is successful, and q = a number of adults (sometimes a surprisingly small number) were involved with them. p => q, but q =/> p.
- tgb 6y agoYou're way overthinking this. If I say "it's takes surprisingly little" it's because I'm emphasizing that it does not take a lot and instead a small amount is sufficient. Saying "it takes a surprisingly small number of emotionally stable adults to help kids " is emphasizing that it does not require a large number of stable adults and therefore a small number is sufficient. How many does it take? Surprisingly few. A few suffice. How else can you read this? You're surprised that a necessary number is small? But it's always necessary to have at least 0 stable adults and that's as small as possible - you can't get surprised that it's small! Instead you must be surprised that the sufficient number is small.
- lostcolony 6y agoNow who is overthinking it? :P I'm not making any comment on the number of adults; I'm just responding to the original argument that "well, I know foster kids who haven't been successful, who had supportive adults", and the ensuing argument, to further explain why that isn't a negation.