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GentlePurpleRain
GentlePurpleRain
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"Thus we've proved that if any set of n−1$n−1$ Canadians have the same age, any set of n$n$ Canadians have the same age." No, you've actually proven the opposite: if a set of n$n$ Canadians have the same age, then any subset of n−1$n−1$ Canadians have the same age. More precisely, you've proven that any n-1$n-1$ subset is comprised of members the same age as members of any other n-1$n-1$ subset. You've restricted yourself to subsets of size n-1$n-1$, but it should be possible to extend the logic for subsets of any size.

My first thought was that you can't work backwards to size n=1$n=1$, because your logic depends on having two sets of at least two people. However, you can work backwards to n=2$n=2$ or 3$n=3$, and I'm pretty sure you can find a set of 2$2$ or 3$3$ people in Canada who do, in fact, have the same age.

"Thus we've proved that if any set of n−1 Canadians have the same age, any set of n Canadians have the same age." No, you've actually proven the opposite: if a set of n Canadians have the same age, then any subset of n−1 Canadians have the same age. More precisely, you've proven that any n-1 subset is comprised of members the same age as members of any other n-1 subset. You've restricted yourself to subsets of size n-1, but it should be possible to extend the logic for subsets of any size.

My first thought was that you can't work backwards to size n=1, because your logic depends on having two sets of at least two people. However, you can work backwards to n=2 or 3, and I'm pretty sure you can find a set of 2 or 3 people in Canada who do, in fact, have the same age.

"Thus we've proved that if any set of $n−1$ Canadians have the same age, any set of $n$ Canadians have the same age." No, you've actually proven the opposite: if a set of $n$ Canadians have the same age, then any subset of $n−1$ Canadians have the same age. More precisely, you've proven that any $n-1$ subset is comprised of members the same age as members of any other $n-1$ subset. You've restricted yourself to subsets of size $n-1$, but it should be possible to extend the logic for subsets of any size.

My first thought was that you can't work backwards to size $n=1$, because your logic depends on having two sets of at least two people. However, you can work backwards to $n=2$ or $n=3$, and I'm pretty sure you can find a set of $2$ or $3$ people in Canada who do, in fact, have the same age.

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Florian F
Florian F
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"Thus we've proved that if any set of n−1 Canadians have the same age, any set of n Canadians have the same age." No, you've actually proven the opposite: if a set of n Canadians have the same age, then any subset of n−1 Canadians have the same age. More precisely, you've proven that any n-1 subset is comprised of members the same age as members of any other n-1 subset. You've restricted yourself to subsets of size n-1, but it should be possible to extend the logic for subsets of any size.

"Thus we've proved that if any set of n−1 Canadians have the same age, any set of n Canadians have the same age." No, you've actually proven the opposite: if a set of n Canadians have the same age, then any subset of n−1 Canadians have the same age. More precisely, you've proven that any n-1 subset is comprised of members the same age as members of any other n-1 subset. You've restricted yourself to subsets of size n-1, but it should be possible to extend the logic for subsets of any size.

My first thought was that you can't work backwards to size n=1, because your logic depends on having two sets of at least two people. However, you can work backwards to n=2 or 3, and I'm pretty sure you can find a set of 2 or 3 people in Canada who do, in fact, have the same age.

"Thus we've proved that if any set of n−1 Canadians have the same age, any set of n Canadians have the same age." No, you've actually proven the opposite: if a set of n Canadians have the same age, then any subset of n−1 Canadians have the same age. More precisely, you've proven that any n-1 subset is comprised of members the same age as members of any other n-1 subset. You've restricted yourself to subsets of size n-1, but it should be possible to extend the logic for subsets of any size.

My first thought was that you can't work backwards to size n=1, because your logic depends on having two sets of at least two people. However, you can work backwards to n=2 or 3, and I'm pretty sure you can find a set of 2 or 3 people in Canada who do, in fact, have the same age.

"Thus we've proved that if any set of n−1 Canadians have the same age, any set of n Canadians have the same age." No, you've actually proven the opposite: if a set of n Canadians have the same age, then any subset of n−1 Canadians have the same age. More precisely, you've proven that any n-1 subset is comprised of members the same age as members of any other n-1 subset. You've restricted yourself to subsets of size n-1, but it should be possible to extend the logic for subsets of any size.

My first thought was that you can't work backwards to size n=1, because your logic depends on having two sets of at least two people. However, you can work backwards to n=2 or 3, and I'm pretty sure you can find a set of 2 or 3 people in Canada who do, in fact, have the same age.

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anonymous coward
anonymous coward
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"Thus we've proved that if any set of n−1 Canadians have the same age, any set of n Canadians have the same age." No, you've actually proven the opposite: if a set of n Canadians have the same age, then any subset of n−1 Canadians have the same age. More precisely, you've proven that any n-1 subset is comprised of members the same age as members of any other n-1 subset. You've restricted yourself to subsets of size n-1, but it should be possible to extend the logic for subsets of any size.

My first thought was that you can't work backwards to size n=1, because your logic depends on having two sets of at least two people. However, you can work backwards to n=2 or 3, and I'm pretty sure you can find a set of 2 or 3 people in Canada who do, in fact, have the same age.