I, like some of my esteemed peers and colleagues in the Mathematics world, have a penchant for pedantry (never thought i'd ever use those two words in the same sentence).
I don't do it for the sake of just 'doing it' (if so, is it still pedantry?); rather, I find that sometimes a little error, can actually throw you off and prevent some of the more subtle understanding that is needed in mathematics; especially in pure mathematics.
I was reading Unit AA1 of M208, Pure Mathematics, and spotted what I believe might be an error; but I am second guessing myself, as to whether it is just my limited understanding of the subject.
The possible errata I refer to, is on page 28, Example 4.3. In the solution it describes the inequality \[1 - {\textstyle{1 \over {{n^2}}}} \le 1\] for n = 1,2,....
However, I would have thought that as n tends to infinity, that \[1 - {\textstyle{1 \over {{n^2}}}}\] would approach 1 but never actually reach it. So, therefore, that inequality should be just \[1 - {\textstyle{1 \over {{n^2}}}} < 1\]
Do I have that right? I know it is a small thing, but I really must make sure, that my understanding is solid.
ps: if the latex above is not displaying, do let me know. I am testing out Mathtype 6.7.
This has been discussed in the M208 forum. The definition requires you to show 1-1/n^2 <= M. It is true that 1-1/n^2 < 1 from which it follows that the weaker condition 1-1/n^2<=1 is also true.
ReplyDeleteRemember that <= has an 'or' in it so the statement 1 <= 2 is true as it says (1 < 2) or (1 = 2) and the first part is true.
You could also think of <= as meaning 'not greater than'.
PS The maths output shows in Firefox but not in IE.
Stevem
Hi
ReplyDeleteI just fished out the AA1 text, and looked at this bit. I think that with the context, it is fine as ≤. I say this because, the strategy says that to show that M is the least upper bound you need to check that x is ≤ M for all x in the set E. Well, whether or not the expression reaches 1, it is still in the collection of all things less than or equal it, in my opinion. Similarly, I don't suppose it's incorrect really to say ≤ 2 for that expression. All values are indeed either less than or equal 2, it just happens that there are none of the latter. But that's irrelevant here as we're supposed to be seeking the least upper bound, and that aint 2; I was just making a point.
The reason, I guess, that the ≤ is used in the strategy is a question of generality. If the least upper bound is within the set in question, take [1, 3] for example, it will be that all numbers in the set are less than or equal to M (=3). Whereas if the least upper bound is not, like in (1, 3), they'll all be just plain less than. But rather than say, "well, if it's in the set, use this condition, and if it's not, use this other condition", why not generalise and say that in both cases it is ≤?
By the way, if it were a mistake, would not the solution to Example 4.2 above also be wrong? Because it says x ≤ 2 for all x in [0, 2).
Anyway, that's just my opinion on it; I haven't looked at this textbook in two years, and the big pencil hole running through from the front to back cover would suggest that I had some issues at the time.
All the best,
Neil H
It's a good question, the answer to which I have seen fudged many times.
ReplyDeleteHere's how I think of it, which may help (or may not):
Suppose you have a function which approaches 0 from the rhs:
y=x, for example.
As x gets smaller and approaches 0, so y does too, in a linear fashion, the graph of the resulting function obviously being the line y=x.
So, when x = 0, y = 0. Obvious.
Now, turn x into a fraction ie x/1.
Nothing has changed.
We still get y = 0 at x = 0.
Now multiply top and bottom of x/1 by x, to give you an "equivalent" fraction.
We now have x-squared/x is equivalent to x/1.
Except, we now have a problem at 0, because the denominator becomes 0, making division at this point undefined.
So two equivalent fractions don't produce identical graphs - one has a hole in it at 0.
How to make the puncture disappear? Define the limit as x approaches 0 of x-squared/x to be zero.
We can get as close to 0 as we like by making x smaller and smaller until, "in the limit", we have zero.
Now take the argument out to infinity in your case and 1/n-squared is zero in the limit, producing 1-0=1.
1 is clearly not less than 1, hence the book's definition.
I am afraid not. However, I was also caught out by this and there are plenty of threads about it on the OU M208 forum. The less than or equal to inequality is still true because of the "or". So, for example, 0 is less than or equal to 1 because the "is less than" bit is true. Hope this makes sense.
ReplyDeleteDuncan.
Thanks for the responses. I always think that it is a good exercise, in and of itself, to challenge or probe the learning material. I find so much learning coming from it. I think that is what is very exciting about mathematics, despite its rigour; it still allows for stimulating discussion and thought provoking musings.
ReplyDelete