Inequalities on the Gridiron
Q: Why are Buffalo Bills unlike Dollar Bills? A: Dollar Bills are good for 4 quarters
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| Reddit “outsmarting math” source |
Josh Allen is not appearing in today’s Super Bowl. He led the Buffalo Bills to not just one but two go-ahead touchdowns in the final 2:00 of the game at Kansas City three weeks ago, but the Bills lost both leads and KC won in overtime.
Today we note an inequality that treats the real numbers like a football gridiron.
The two-minute drill is the hallmark of quarterback heroism. KC’s Patrick Mahomes, however, demonstrated the 13-second drill in two plays plus a tying field goal. That is, the Bills were good for 3 quarters plus 14:47. But Mahomes is not in the Super Bowl either. Two weeks ago, he lost the magic touch in both the closing minute of the fourth quarter and the beginning of overtime as KC squandered a big lead and lost to the Cincinnati Bengals. Whose quarterback, Joe Burrow, is playing in today’s Super Bowl, opposite Matthew Stafford of the Los Angeles Rams.
The American football field is divided into 100 yards, but rarely subdivided beyond that. The announcers may refer to “a long two yards” or “a short 3,” but never 2.5 yards. It is not just the announcers. Official statistics are kept in units of whole yards, more often rounded up than down. A fourth-down quarterback plunge with 3 inches to go still counts as a 1-yard gain. Perhaps it is essential to the yard that it not be divided into dyadic or decimal units, the way the meter is. As the US celebrates an event still enjoyed by “one nation indivisible,” we note the indivisible.
Inequalities
Godfrey Hardy, John Littlewood, George Polya are famous for many things separately and together. All three lent their names to the timeless book Inequalities. It codifies the theory of real inequalities.
An Inequality
We recently had reason to look at the following inequality:
If
, then
We noted that this fails in general: Let 
This is not even close. But wait—the following lemma is true:
Lemma 1 If
provided
are integers.
The proof is quite simple. We note that
But this shows that 
This proves the inequality.
Open Problems
Is this interesting at all? We have an application of the above lemma, which we will discuss in the future. Are there other good examples of inequalities that are general and natural and only hold over the integers? Or is seeking them like trying to “outsmart math itself”?
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| SB Nation source |
Hi Dick,
I hope you are fine! Look at this tiny change:
Statement 2: If a^2 – b^2 = c > 0, then c >= a + b provided a and c are integers (b could be either an integer or not).
Now, we use different variables with the same label (a, b, c) in this NP-complete problem:
QUADRATIC CONGRUENCES: Given the positive integers a, b, and c, is there a positive integer x 0 where d = bk (multiple of b, that is, k is a natural number).
This is the same as (x + a^0.5)(x – a^0.5) = d.
If the Statement 2 is true, then d >= x + a^0.5.
Statement 3: We can say the triple (a, b, c) does not belong to QUADRATIC CONGRUENCES when there is not any multiple of b (the number d) such that
2 * a^0.5 < d < c + a^0.5.
Algorithm for the Statement 3: we can check in polynomial time whether there is not any multiple of b (the number d) such that 2 * a^0.5 < d < c + a^0.5 just checking whether
b * floor((c + a^0.5)/b) <= .2 * a^0.5
where floor(..) is the floor function.
I don't know how clever or simple could sound the Statement 2 and 3, but I think this is an evident way to use the inequality that you mentioned in this post.,
Regards…
I saw my previous comment was cut. Here it is the part that was deleted:
QUADRATIC CONGRUENCES: Given the positive integers a, b, and c, is there a positive integer x 0 where d = bk (multiple of b, that is, k is a natural number). This is the same as (x + a^0.5)(x – a^0.5) = d.
Hi Dick,
I hope you are fine! Look at this tiny change:
Statement 2: If a^2 – b^2 = c > 0, then c >= a + b provided a and c are integers (b could be either an integer or not).
Now, we use different variables with the same label (a, b, c) in this NP-complete problem:
QUADRATIC CONGRUENCES: Given the positive integers a, b, and c, is there a positive integer x 0 where d = bk (multiple of b, that is, k is a natural number). This is the same as (x + a^0.5)(x – a^0.5) = d.
If the Statement 2 is true, then d >= x + a^0.5.
Statement 3: We can say the triple (a, b, c) does not belong to QUADRATIC CONGRUENCES when there is not any multiple of b (the number d) such that
2 * a^0.5 < d < c + a^0.5.
Algorithm for the Statement 3: we can check in polynomial time whether there is not any multiple of b (the number d) such that 2 * a^0.5 < d < c + a^0.5 just checking whether
b * floor((c + a^0.5)/b) <= .2 * a^0.5
where floor(..) is the floor function.
I don't know how clever or simple could sound the Statement 2 and 3, but I think this is an evident way to use the inequality that you mentioned in this post.,
Regards…
I don’t understand why the message is cut. I will use it in another way:
QUADRATIC CONGRUENCES: Given the positive integers a, b, and c, is there a positive integer x is less than c such that (x^2 mod b) = a? This is equivalent to x^2 – a = d that is greater than 0 where d = bk (multiple of b, that is, k is a natural number). This the same as (x + a^0.5)(x – a^0.5) = d.
Dick, you could delete the second and third comment above (and this fifth too). Sorry for this bunches of comments, but a “less than” symbol was splitting my messages in a messy way.
The correctness of the inequality has got nothing to do with integers or non-integers:
For non-negative $a$ and $b$, the equality $a^2-b^2=c>0$ implies the inequality $c\ge a+b$ if and only if $a-b\ge1$.
As you surely know, inequalities that are true for integers but not for reals form the basis for cutting planes in integer programming. For example, suppose a, b, c, d, e all lie between 0 and 1 inclusive, and a+b, b+c, c+d, d+e, and e+a are each less than or equal to 2. If all five numbers have to be integers, then a+b+c+d+e is at most 2, but if they can be real numbers, then you can set all five variables equal to 1/2.
Also, most proofs of irrationality or transcendence reduce to the fact that there is no integer strictly between 0 and 1. But such results may not be in the spirit of what you’re asking for.
Sorry, of course I meant to say that a+b, b+c, c+d, d+e, and e+a are each less than or equal to 1.