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I am working on the following problem from Rustan Leino's website (http://leino.science/puzzles/the-exact-batting-average/):

At some point during a baseball season, a player has a batting average of less than 80%. Later during the season, his average exceeds 80%. Prove that at some point, his batting average was exactly 80%.

I have what I think is the beginning of a proof, but I was hoping to get some feedback on whether my solution is on the right track.

First, I want to point out a critical assumption. By a player having a batting average less than 80%, I take the puzzle to mean the player has a batting average less than 80% after having had at least 1 at bat.

Suppose the batting average does not equal .8 at any point. We can show this is a contradiction (I think).

Observe that if the batting average is below .8, eventually exceeds .8, and never equals .8, then there will eventually be some hit such that before the hit the batting average is below .8 and afterwards the batting average is above .8. Prior to this hit we have $a/b<4/5$ where $a$ is the total number of hits and $b$ is the total number of at bats; after this hit, we have $(a+1)/(b+1)>4/5$ where $a$ is an integer $\geq0$ and $b$ is an integer $\geq1$. This implies the following two inequalities:

(1) $5a<4b$,and

(2) $5a+1>4b$.

I think these two inequalities produce a contradiction. If $5a$ is an integer that is less that $4b$, adding $1$ should at most make $5a$ equal to $4b$ since $5a\leq4b-1$.

Am I on the right track here?

Yonas Oberlin
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    makes sense to me – gt6989b Jul 13 '21 at 03:29
  • @gt6989b Yay! That's encouraging. – Yonas Oberlin Jul 13 '21 at 05:38
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    I would consider this "10cm before the finish line" in a 100m race, rather than "on the right track". – Ingix Jul 13 '21 at 09:42
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    We can guess that $a$ is hits and $b$ is at-bats, but I would state it explicitly before using $a$ or $b.$ – David K Jul 13 '21 at 16:49
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    Conditions (1) and (2) are contradictory for any integers $a$ and $b,$ so you don't need the "greater than or equal to $1$" condition after that point. I suppose it's worth assuming that $b \geq 1$; we must assume that a player has to have at least one at-bat to have a batting average below $.800$, otherwise a player who hit on their first at-bat would be a counterexample to the theorem. But a player can have an average below $.800$ with no hits, and the assumption $a\geq1$ means technically you'd need to treat no hits as a special case. – David K Jul 13 '21 at 17:10
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    I see now you added definitions of $a$ and $b.$ Nicely done! – David K Jul 13 '21 at 17:11
  • @DavidK Right, I should definitely account for a = 0. Thanks very much for pointing this out! – Yonas Oberlin Jul 13 '21 at 17:25

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Yes, it looks like you're on the right track.

We have $5a<4b$ and $5a+1>4b$, which means

$$5a<4b<5a+1 \implies 0<4b-5a<1$$

We know that $a$ is a non-negative integer and $b$ is a natural number, both by definition. In other words, $a$ and $b$ are both integers.

The integers are closed under multiplication, so $4b$ and $5a$ must both be integers. The integers are also closed under subtraction, so $4b-5a$ is also an integer.

At this point, you would have to prove that no integer exists between $0$ and $1$. From what I've seen in browsing for proofs of this, they mostly involve the well-ordering principle, as shown in this question here. I'm not sure how deep the question wants you to go though: maybe the prior paragraph would be enough to establish a contradiction.

Kman3
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  • The last point seems very interesting. I think I'll go off and see if I can prove this. – Yonas Oberlin Jul 13 '21 at 05:49
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    I don't agree that you have to prove that no integer exists between $0$ and $1$. This is not the type of problem for which going right back to foundational principles is appropriate. – TonyK Jul 13 '21 at 17:33
  • @TonyK As I stated in my answer, I wasn't sure what kind of problem this was and how deep you would have to go. I added that paragraph for the sake of completeness. I will let OP judge whether that part is required or not based on their knowledge of the problem. – Kman3 Jul 13 '21 at 19:32
  • And your spiel about "closed under multiplication", "closed under subtraction": that is jargon for the sake of it, more likely to confuse than enlighten. Just say "$a$ and $b$ are integers, so $4b,5a,$ and $4b-5a$ are also integers." – TonyK Jul 13 '21 at 19:44
  • @TonyK I've stated this already, and I will repeat it again here: I cannot judge based on the problem how much proof is sufficient. There's no "jargon" here; it depends on how in-depth the questioner wants to go. Of course if $b$ is an integer then $4b$ is an integer, but that statement may require proof depending on how the problem is laid out. It seems you're objecting not because I solve the problem in a way that's incorrect, but in a way that you don't like. The questioner can solve it however they want. If they want to stop at $0<4b-5a<1$, great. If not, keep going. What's the issue? – Kman3 Jul 13 '21 at 20:50
  • The issue is that you have taken a simple problem about positive integers and turned it into a disquisition on the foundations of mathematics. How far down do you want to go? Are you suggesting that we start with $0={},1={0},2={1}$ etc? Or what? It's like you're just showing off. – TonyK Jul 13 '21 at 21:28
  • @TonyK You clearly haven't read what I've said to you, so there's no point in continuing this discussion. The level of rigor depends on what the problem demands. Can $4b$ can both be an integer and lie between two consecutive integers? Of course not. If you feel no supplemental reasoning is necessary, that's fine. I'm not going to stop you. However, be prepared for it to be asked of you. I did not suggest anything you describe. If you don't like my answer, add your own. Otherwise, quit projecting your disdain for mathematical rigor onto me and go pick a fight with somebody else. – Kman3 Jul 13 '21 at 22:28