Talk:Intermediate value theorem

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I'm not a mathematician, but there's a bit of this proof that seems wrong. It goes

"Suppose first that f (c) > u."

I don't see how we can suppose this since by definition

    c = sup {x in [a, b] : f(x) ≤ u}.

which I understand as meaning that c is the largest number of those real numbers x such that f(x) is less or equal to u. If so, how could f(c) be greater than u??

Wrong. sup does not mean the largest member of a set, but rather the smallest upper bound of the set, i.e., the smallest number that no member of the set exceeds. If the set has a largest member, then that is the set's sup, but the set of all numbers strictly less than 10 has no largest member, but still has a sup, which is 10. Michael Hardy 01:32, 4 Jan 2004 (UTC)

My mistakem, which I realised on Friday evening when I looked up the definition of supremum. Wondering why I didn't get this from Wikipedia, I looked at the definition there supremum, and saw a non-mathematical definition precedes it, which could cause confusion. DB.

Contents

[edit] Calculus?

The IVT for integration is certainly a result of calculus, but the original IVT involves neither differentiation nor integration but only continuity, so shouldn't it be referred to in the first sentence as a result of analysis? --131.111.249.207 17:31, 2 Jun 2005 (UTC)

I agree, and I changed this

Grokmoo 13:29, 21 October 2005 (UTC)

Hmmm...if it only involves continuity, shouldn't it REALLY be called a result of TOPOLOGY???

[edit] Closed interval

Uncommon though the formation may be, I think the theorem can say f(x) = c for x in [a, b] not just (a, b). [article] says so too. -- Taku 01:07, 13 October 2005 (UTC)

Note that the statement with (a, b) is actually a STRONGER statement. In any case, there's no point in including a and b as endpoints, since it's impossible for them to work (plug them in and try).

It can if you like, but this case it trivial, so I wouldn't worry about it. The standard statement is for (a, b).

[edit] Proof that f(x) = x for some x

The proof right below the intermediate value theorem that there exists some x such that f(x) = x is wrong. This statement is in fact not generally true, even for f(x) continuous. There are additional requirements, for example, if the domain of f is all the reals and the function is bounded (above and below).

I reorganized this article and fixed the proof.

Grokmoo 13:29, 21 October 2005 (UTC)

Would it be useful to add an intuitively understandable example to this page? Something like: If person A is climbing a mountain from 6 to 7 am, and person B is coming down the mountain during the same time interval, then there has to be some time t in that time interval when they are both at exactly the same altitude?

The example of the "without lifting the pencil" is quite intuitive and very nice.

I agree. However, opening the page and being faced with a set of formulae isn't very nice. It would be better if someone put the following paragraph at the beginning, or in the introduction:

""This captures an intuitive property of continuous functions: given f continuous on [1, 2], if f (1) = 3 and f (2) = 5 then f must be equal to 4 somewhere between 1 and 2. It represents the idea that the graph of a continuous function can be drawn without lifting your pencil from the paper.""

Khedron 00:41, 23 December 2006 (UTC)

Too bad that this idea of "not lifting the pen from the paper" is quite wrong. For example, consider the Vitali Cantor Function. It is a continuous (even holder continuous) function but it is quite impossible to draw (or even, to think of). I think that this kind of "generally true" statements should be avoided when writing mathematics. Francesco di Plinio (francesco.diplinio@libero.it) —Preceding unsigned comment added by 151.57.122.59 (talk) 16:53, 27 September 2007 (UTC)

[edit] left and right neighbourhood

"Suppose first that f (c) > u ... whenever | x - c | < δ" I think it should be c - δ < x < c (left neighbourhood) and c < x < c + δ (right neighbourhood) for f(c) < u so we can omit absolut function. In first whenever c - δ < x < c --> f(x) - f(c) < 0 so |f(x) - f(c)| = -f(x) + f(c) —The preceding unsigned comment was added by 149.156.124.14 (talk) 15:48, 23 January 2007 (UTC).