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LostInParadise's avatar

Is this intuitively obvious?

Asked by LostInParadise (29137points) February 8th, 2021
9 responses
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Suppose that you are looking at information about 21 people on your computer screen. It is currently displaying the first 20. You look through the information and notice that everyone is from a different city. You scroll the list down and notice that the second to twenty-first people are from different cities. Is it obvious that everyone would be from a different city provided that the first and twenty-first were? How would you convince someone who was uncertain?

This is the best that I could do. Saying everyone is from a different city is the same as saying every pair is from a different city. From the first view we can conclude that every pair of people, not including the 21st person, is from a different city. From the second view, we can add every pair that includes the 21st person, except for the first person. The only pair that is missing is the 1st and 21st. If they are from different cities then everyone is.

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doyendroll's avatar

Can you play snap?

LuckyGuy's avatar

I intuitively thought it is possible to have #1 and #21 be from the same city and still satisfy your statements. Are there more than 21 cities?
I need to think more about why that is not true.

Zaku's avatar

It is intuitively obvious to me. But I’ve also done quite a few logical proofs, was raised by analytical parents, and I’ve programmed computers since I was about 11 years old, and I’d either just let them be ignorant, or toss them a diagram or a few lines of logic.

The first one doesn’t match the middle 19.
The last one doesn’t match the middle 19.
The first one doesn’t match the bottom one.

Yeah, that should be more than enough. If it’s not clear, they’ve got a logic problem, so actually if I cared, I’d ask them what they were thinking, or for an example where the above would be true but they wouldn’t all be unique. And/or ask them if they can see that the middle 19 might as well be 1 for this purpose, so it’s like A != B != C != A.

LostInParadise's avatar

Be careful with your !=‘s. A != B != C does not imply A != C. Not equals is a little trickier to work with. If I had said that the first 20 were the same and then said 2 to 21 were all the same then obviously they would all be the same.

Zaku's avatar

@LostInParadise Yeah, I wasn’t clear what I meant by what I wrote.

Your problem description however, once we accept “not matching any of the middle 19” may as well be represented by != B, can be reduced to:

1. A != B
2. B != C
3. A != C

And that actually describes a logical triangle. No further steps are needed past what you told us about the situation, because you told us all three significant types of possible comparison were not equal.

So the confusion probably would lie in not understanding that the fact that B is a set does not actually add any complexity versus it being one value.

A Venn diagram might make that immediately clear to more visual people.

Zaku's avatar

My guess is that @LuckyGuy ‘s listening just tuned out by the time he read the part that you actually told him #1 was not the same as #21, or perhaps more likely, he didn’t get what you meant by the crucial words: “provided that the first and twenty-first were”.

doyendroll's avatar

Yes, it is obvious.

flutherother's avatar

1 – 20 are all different. Now think of 21 as a single case. We are told 21 is different from 2 – 20 and we are also told it is different from 1. Looked at this way it is intuitively obvious that 21 must be different from the others.

LuckyGuy's avatar

@Zaku Yep! You got me. I missed the “provided….” Ugh! I should read more carefully!

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