Fun with Math
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18-01-2014, 12:12 AM
RE: Fun with Math
(18-01-2014 12:09 AM)Tartarus Sauce Wrote:  
(18-01-2014 12:03 AM)GirlyMan Wrote:  Spot the fallacy. This one's pretty hard to spot. I didn't spot it. You gotta know something about proof by induction to follow it.

1) In any group that consists of just one person, everybody in the group has the same age, because after all there is only one person!
2) Therefore, statement S(1) is true.
3) The next stage in the induction argument is to prove that, whenever S(n) is true for one number (say n=k), it is also true for the next number (that is, n = k+1).
4) We can do this by (1) assuming that, in every group of k people, everyone has the same age; then (2) deducing from it that, in every group of k+1 people, everyone has the same age.
5) Let G be an arbitrary group of k+1 people; we just need to show that every member of G has the same age.
6) To do this, we just need to show that, if P and Q are any members of G, then they have the same age.
7) Consider everybody in G except P. These people form a group of k people, so they must all have the same age (since we are assuming that, in any group of k people, everyone has the same age).
8) Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.
9) Let R be someone else in G other than P or Q.
10) Since Q and R each belong to the group considered in step 7, they are the same age.
11) Since P and R each belong to the group considered in step 8, they are the same age.
12) Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.
13) We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.
14) The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n.

Jelly Donut≠Fire Hydrant.

Did I get it right?

That is one of the correct answers. Thumbsup

There is only one really serious philosophical question, and that is suicide. -Camus
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18-01-2014, 12:23 AM
RE: Fun with Math
(17-01-2014 11:53 PM)GirlyMan Wrote:  
(17-01-2014 09:53 PM)Hafnof Wrote:  ... and this result is necessary to a proper understanding of quantum physics(!)





Some crazy fucking shit ain't it, Hafnof?

I got lost at the multiplication part!!!

mathS

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18-01-2014, 12:26 AM
RE: Fun with Math
(17-01-2014 04:49 PM)GirlyMan Wrote:  Spot the fallacy.

1) Let a=b.
2) a^2=ab
3) a^2 + a^2 = a^2 + ab
4) 2a^2 = a^2 + ab
5) 2a^2 - 2ab = a^2 + ab - 2ab
6) 2a^2 - 2ab = a^2 - ab
7) 2(a^2 - ab) = a^2 - ab
8) 2(a^2 - ab)/(a^2 - ab)=(a^2-ab)/(a^2-ab)
9) 2 = 1 QED

a^2 - ab = 0. You can't divide by 0 silly.


You see stuff in numbers... I see stuff in colours and light. I get freakin' ECSTATIC with perfect light.
Is that what these numbers are? Beautiful universal language, for sure. I can really appreciate that.
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18-01-2014, 12:32 AM (This post was last modified: 18-01-2014 12:49 AM by GirlyMan.)
RE: Fun with Math
(18-01-2014 12:26 AM)LadyJane Wrote:  You see stuff in numbers... I see stuff in colours and light. I get freakin' ECSTATIC with perfect light.
Is that what these numbers are? Beautiful universal language, for sure. I can really appreciate that.

Girly thinks I might kinda wanna try me some of that ecstasy color shit you got ... Thumbsup




There is only one really serious philosophical question, and that is suicide. -Camus
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18-01-2014, 12:57 AM
RE: Fun with Math
(18-01-2014 12:03 AM)GirlyMan Wrote:  Spot the fallacy. This one's pretty hard to spot. I didn't spot it. You gotta know something about proof by induction to follow it.

Everyone's the same age.

1) In any group that consists of just one person, everybody in the group has the same age, because after all there is only one person!
2) Therefore, statement S(1) is true.
3) The next stage in the induction argument is to prove that, whenever S(n) is true for one number (say n=k), it is also true for the next number (that is, n = k+1).
4) We can do this by (1) assuming that, in every group of k people, everyone has the same age; then (2) deducing from it that, in every group of k+1 people, everyone has the same age.
5) Let G be an arbitrary group of k+1 people; we just need to show that every member of G has the same age.
6) To do this, we just need to show that, if P and Q are any members of G, then they have the same age.
7) Consider everybody in G except P. These people form a group of k people, so they must all have the same age (since we are assuming that, in any group of k people, everyone has the same age).
8) Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.
9) Let R be someone else in G other than P or Q.
10) Since Q and R each belong to the group considered in step 7, they are the same age.
11) Since P and R each belong to the group considered in step 8, they are the same age.
12) Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.
13) We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.

14) The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n... when and only when we make the assumption that in our group of k people, everyone has the same age.

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18-01-2014, 01:08 AM
RE: Fun with Math
(18-01-2014 12:32 AM)GirlyMan Wrote:  
(18-01-2014 12:26 AM)LadyJane Wrote:  You see stuff in numbers... I see stuff in colours and light. I get freakin' ECSTATIC with perfect light.
Is that what these numbers are? Beautiful universal language, for sure. I can really appreciate that.

Girly thinks I might kinda wanna try me some of that ecstasy color shit you got ... Thumbsup




I've never seen this, I'm laughing. I'm thinking all your fancy number combos are what dewy's after. I'm more like this (ironically).



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18-01-2014, 01:15 AM
RE: Fun with Math
(18-01-2014 12:57 AM)DLJ Wrote:  
(18-01-2014 12:03 AM)GirlyMan Wrote:  Spot the fallacy. This one's pretty hard to spot. I didn't spot it. You gotta know something about proof by induction to follow it.

Everyone's the same age.

1) In any group that consists of just one person, everybody in the group has the same age, because after all there is only one person!
2) Therefore, statement S(1) is true.
3) The next stage in the induction argument is to prove that, whenever S(n) is true for one number (say n=k), it is also true for the next number (that is, n = k+1).
4) We can do this by (1) assuming that, in every group of k people, everyone has the same age; then (2) deducing from it that, in every group of k+1 people, everyone has the same age.
5) Let G be an arbitrary group of k+1 people; we just need to show that every member of G has the same age.
6) To do this, we just need to show that, if P and Q are any members of G, then they have the same age.
7) Consider everybody in G except P. These people form a group of k people, so they must all have the same age (since we are assuming that, in any group of k people, everyone has the same age).
8) Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.
9) Let R be someone else in G other than P or Q.
10) Since Q and R each belong to the group considered in step 7, they are the same age.
11) Since P and R each belong to the group considered in step 8, they are the same age.
12) Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.
13) We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.

14) The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n... when and only when we make the assumption that in our group of k people, everyone has the same age.

Yup. That's the gist of it. Step 9 is the problem, may not be anybody else in G but P & Q. Well done DLJ.

There is only one really serious philosophical question, and that is suicide. -Camus
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18-01-2014, 01:25 AM
RE: Fun with Math
(18-01-2014 01:15 AM)GirlyMan Wrote:  
(18-01-2014 12:57 AM)DLJ Wrote:  14) The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n... when and only when we make the assumption that in our group of k people, everyone has the same age.

Yup. That's the gist of it. Step 9 is the problem, may not be anybody else in G but P & Q. Well done DLJ.


Lucky guess.

For any further advice you require, consulting day rates ($2,000) apply.

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18-01-2014, 01:35 AM (This post was last modified: 18-01-2014 01:42 AM by GirlyMan.)
RE: Fun with Math
(18-01-2014 01:25 AM)DLJ Wrote:  
(18-01-2014 01:15 AM)GirlyMan Wrote:  Yup. That's the gist of it. Step 9 is the problem, may not be anybody else in G but P & Q. Well done DLJ.


Lucky guess.

For any further advice you require, consulting day rates ($2,000) apply.





Dance motherfucker, dance.

There is only one really serious philosophical question, and that is suicide. -Camus
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18-01-2014, 06:48 AM (This post was last modified: 18-01-2014 06:55 AM by Chas.)
RE: Fun with Math
(18-01-2014 12:03 AM)GirlyMan Wrote:  Spot the fallacy. This one's pretty hard to spot. I didn't spot it. You gotta know something about proof by induction to follow it.

Everyone's the same age.

1) In any group that consists of just one person, everybody in the group has the same age, because after all there is only one person!
2) Therefore, statement S(1) is true.
3) The next stage in the induction argument is to prove that, whenever S(n) is true for one number (say n=k), it is also true for the next number (that is, n = k+1).
4) We can do this by (1) assuming that, in every group of k people, everyone has the same age; then (2) deducing from it that, in every group of k+1 people, everyone has the same age.
5) Let G be an arbitrary group of k+1 people; we just need to show that every member of G has the same age.
6) To do this, we just need to show that, if P and Q are any members of G, then they have the same age.
7) Consider everybody in G except P. These people form a group of k people, so they must all have the same age (since we are assuming that, in any group of k people, everyone has the same age).
8) Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.
9) Let R be someone else in G other than P or Q.
10) Since Q and R each belong to the group considered in step 7, they are the same age.
11) Since P and R each belong to the group considered in step 8, they are the same age.
12) Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.
13) We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.
14) The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n.

Step 4 assumes the conclusion. It is invalidly applying a universal quantifier, going from 'there exists an x' to 'for all x'.

But I'm not even half way through my first cup of coffee, so I may be wrong. Drinking Beverage

Skepticism is not a position; it is an approach to claims.
Science is not a subject, but a method.
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