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chowching
11-25-2014, 12:19 PM
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!

md2000
11-25-2014, 12:31 PM
a+b≥x - true

a+b≥x−1 rewritten as:
a+b+1≥x

I don't see a contradiction, except that the second set allows for a+b to be 1 less than the solution to the first.

So, for x=0 the first solution is any a+b≥0 but the second is any a+b≥-1
The first solution is a subset of the second?

Thudlow Boink
11-25-2014, 12:32 PM
Assuming I'm understanding you correctly:

a+b≥x-1 doesn't imply that a+b is ever necessarily equal to, or even close to, x-1. In fact, given a+b≥x it automatically follows that a+b≥x-1.

If you could show that a+b was ever equal to a number less than x, such as x-1, you would have your contradiction.

robert_columbia
11-25-2014, 12:34 PM
Actually, Jones v. Smith established a presumption that a+b<x, but this precedent seems to apply only in Ohio.

ultrafilter
11-25-2014, 01:14 PM
5 > 2.

Johnny L.A.
11-25-2014, 01:30 PM
Actually, Jones v. Smith established a presumption that a+b<x, but this precedent seems to apply only in Ohio.

Care to save us some googling?

sbunny8
11-25-2014, 02:22 PM
Actually, Jones v. Smith established a presumption that a+b<x, but this precedent seems to apply only in Ohio.
Care to save us some googling?

I think he's making a joke about the urban legend (http://www.snopes.com/religion/pi.asp) 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 (http://en.wikipedia.org/wiki/Indiana_Pi_Bill#In_popular_culture) (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.

sbunny8
11-25-2014, 02:29 PM
Is my argument correct? Did I arrive at a contradiction?

Short answer: No, you didn't.

But, if it had led you to a specific solution where a+b is between x and x-1 then THAT would be a contradiction.

DrCube
11-25-2014, 02:39 PM
"x" is greater than or equal to "x - 1" so "a + b" can certainly be both >= x AND >= x-1. There is no contradiction.

Senegoid
11-25-2014, 03:04 PM
5 > 2.

But only for extremely large values of 5.

Isilder
11-25-2014, 04:47 PM
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?
!


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.)

chowching
11-25-2014, 07:15 PM
Okay guys here is where I'm coming from.

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.

Thank you.

Indistinguishable
11-25-2014, 08:01 PM
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.
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.

ultrafilter
11-25-2014, 08:15 PM
But then I got that a+b≥x−1. This means that a+b can be x-1.


This is wrong. Again, 5 > 2.

Indistinguishable
11-25-2014, 08:59 PM
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!
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.)

TimeWinder
11-25-2014, 09:31 PM
I guess I have to show then that there really is an instance when a+b=x-1.

I think part of the problem here is that you're using equations for something that's really going to need sets.

I *think* what you're getting at is this:

Hypothesis: for all a and b, a + b >= x

Then, proving that "there exists an a for which there exists a b, such that a+b = x-1" would, indeed, be a contradiction.

Jragon
11-25-2014, 09:43 PM
Suppose there exists some non-negative numbers a,b, and x such that

a + b >= x

Now suppose there's some number

c = b + 1

We should know that:

a + c >= a + b >= x
a + (b + 1) >= x
a + b >= x - 1

This makes it fairly clear that there's no contradiction.

Leo Bloom
11-26-2014, 12:19 AM
But only for extremely large values of 5.I have seen, and been on the receiving end, of this old physicist's get out of jail trick.

But shouldn't that be "only for extremely large values of 2?"

Senegoid
11-26-2014, 12:54 AM
But shouldn't that be "only for extremely large values of 2?"

You're thinking of that old 2 + 2 assertion.

In the case of 5>2, this would hold for extremely large values of 5 or for extremely small values of 2.

ZenBeam
11-26-2014, 06:55 AM
You're thinking of that old 2 + 2 assertion.

In the case of 5>2, this would hold for extremely large values of 5 or for extremely small values of 2.Seems it could also hold for small values of 5 (for example 3 or 4). Likewise for large values of 2.

Sage Rat
11-26-2014, 07:23 AM
I'm not a mathematician, so this is a question not an answer. But if you have proved the first statement, I would expect it to allow you to say that the second statement is so true, you can even swap the >= sign for a >?

davidm
11-26-2014, 07:49 AM
I think what you're trying to say is (assuming I'm getting the set notation right)

{a+b,...} = {x,...} is contradicted by {a+b,...} = {x-1,...}

If so, then you're correct but using incorrect notation.

Your Great Darsh Face
11-26-2014, 08:32 AM
I'm not a mathematician, so this is a question not an answer. But if you have proved the first statement, I would expect it to allow you to say that the second statement is so true, you can even swap the >= sign for a >?

Indeed you could. But a + b >= x - 1 is still perfectly true even if we have established that the strict equality can never hold.

md2000
11-26-2014, 12:13 PM
a+b≥x - true

a+b≥x−1 rewritten as:
a+b+1≥x

I don't see a contradiction, except that the second set allows for a+b to be 1 less than the solution to the first.

So, for x=0 the first solution is any a+b≥0 but the second is any a+b≥-1
The first solution is a subset of the second?

The solution for a+b≥x is a subset of the solution of a+b≥x−1

Therefore any solution that is true for the first is true for the second, but not everything that is true for the second is true for the first.

All circles are ellipses, but not all ellipses are circles.
All ellipses are conic sections, but not all conic sections are ellipses.

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