So I was proving something and I’m wondering if this line of argument is correct.
Suppose that it is true that given conditions M,N,O; a+b≥x. That is given those conditions the minimum value of a+b is x.
Now I want to prove using contradiction that an argument X is true. So I assume that X is false. Under the assumption that X is false, I arrive at conditions M,N,O. I also have that a+b≥x−1. Which is a contradiction since a+b≥x given conditions M,N,O. Is my argument correct? Did I arrive at a contradiction?
I hope you guys can help. I tried to generalize the problem because it’s more with the reasoning that I am concerned about. Thank you!
I think he’s making a joke about the urban legend which claims that some uneducated part of the US once passed a law saying that pi=3. Various tellings of the legend have it happening in Ohio, Tennessee, or Alabama. There’s a grain of truth in that Indiana once considered a bill which said that pi was a constructable number (which it isn’t) and appeared to imply that pi was equal to 4. The legislators themselves didn’t take the bill seriously, and it was postponed indefinitely.
Don’t know !.. Depends on the methods used to determine the lower bound. And what changed between the two calculations ?
So assume the difference was the method… If BOTH methods (algorithm or formula) are for determining highest lower bounds - that is they are deterministic and accurate rather that being an approximation or producing mere example lower bounds… then you have have a contradiction, and either there was human error, or at least one of the two sets of (M,N,O,method) is wrong .
For the purpose of high school mathematics, when the teacher asks for a lower bound or upper bound, they really want the highest lower bound or lowest upper bound… the BEST answer… the BEST upper or lower bound…
There are occassions that merely know something about a lower bound is enough and there is no need to continue on to determine the highest lower bound. Eg if the lower bound is found to be 1, then you can say that its a +ve value. Which can then say all sorts of stuff, eg that divide by zero never occurs, or that square root is never imaginary… the method for finding a mere lower bound can be trial and error… Guess 1 is a lower bound, merely plug 1 into the equation and test that…
(you can find the highest lower bound by trial and error as long as you avoid approximations.)
Under conditions M,N,O, a+b≥x. This means that the least possible value attainable by a+b is x. The value x can be attained but anymore less than x cannot be. So I might as well say that a+b>x-1. Strictly.
But then I got that a+b≥x−1. This means that a+b can be x-1. Which seems contradictory to me because the inequality above is strict.
I guess I have to show then that there really is an instance when a+b=x-1.
Someone might say “Given conditions M, N, O: a + b ≥ x” to mean “Whenever M, N, and O hold, the value of a + b is at least as large as x, and in fact, it may well always be larger than x”. Someone else (or the same person in a different context) might say “Given conditions M, N, O: a + b ≥ x” to mean “Whenever M, N, and O hold, the value of a + b is at least as large as x. Furthermore, in at least one case, conditions M, N, and O hold, and the value of a + b is exactly equal to x.”
These are two different statements which, alas, can be worded in the same way. Only you can tell us which you actually meant in the argument you were concerned with.
Generalization is a good instinct, and I applaud your having done so. However, I’ll note that it may be easier for us to point the relevant principles out to you if you show us more specifically the argument you were carrying out. (Which is not to diminish the value of abstraction, but only to defer to the pedagogical reality that many people in many cases [of which I suspect this will turn out to be one] find things clearer in concrete examples.)
