Kan dette være rett? (Matematikk med en tvist)

[theJack]

[Oceanborn]

[Mr_Pin]

Det er vel oppnådd enighet om at det ikke er lov å dele på 0 (det blir jo som tidligere nevn tull), dette er jo den klassiske måten å vise at 1=2.
Men nyere matemattikk viser at man kan vise det samme uten denne kjedelige feilen, det hele takket være imaginære tall :
(OBS, jeg finner ikke tegnet for kvadratroten på tastaturet så jeg bruker ^0,5 istedet for (det betyr det samme), husk og at kvadratroten av -1 = i)

Step 1: -1/1 = 1/-1

Step 2: Taking the square root of both sides (-1/1)^0,5 = (1/-1)^0,5

Step 3: Simplifying (-1^0,5)/(1^0,5) = (1^0,5)/(-1^0,5)

Step 4: In other words, i/1 = 1/i.

Step 5: Therefore, i / 2 = 1 / (2i),

Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),

Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

Step 8: (i^2)/2 + 3i/2i = i/2i + 3i/2i

Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

Step 10: and this shows that 1=2.

Bertrand Russell would add, "Since the Pope and I are two, and two equals one, then the Pope and I are one, therefore I am the Pope."

Noen som kan vise at dette er feilaktig?

[knuta]

Den der var stilig :-)
Det tok en stund å se det, selv om jeg fattet mistanke til kvadratroten.

[Mr_Pin]

Stemmer det. *Aplaus*

Problemet ligger i at (1/-1)^0,5 ikke automatisk er lik (1^0,5)/(-1^0,5). Denne forenklingen av kvadratrøtter er helt riktig med reele tall ikke nødvendigvis gyldig når man involverer imaginære tall.
Poenget med denne er selvsagt at (-1*-1)^0,5 ikke er lik (-1)^0,5*(-1)^0,5

Men kan du finne en feil i dette enkle induksjonsbeviset på at alle som bor i Canada er like gamle?

This "proof" will attempt to show that all people in Canada are the same age, by showing by induction that the following statement (which we'll call "S(n)" for short) is true for all natural numbers n:

Step 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!

Step 2: Therefore, statement S(1) is true.

Step 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).

Step 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.

Step 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.

Step 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.

Step 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).

Step 8: Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.

Step 9: Let R be someone else in G other than P or Q.

Step 10: Since Q and R each belong to the group considered in step 7, they are the same age.

Step 11: Since P and R each belong to the group considered in step 8, they are the same age.

Step 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.

Step 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.

Step 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.

[knuta]

[Mr_Pin]

Neida dette er lov.

This step is not the source of the fallacy.
This statement is correctly appealing to the induction assumption.

Remember, in order to prove a statement of the form "if A, then B", what you do is assume A and derive B from it.

In our case, A is the statement S(k): "in any group of k people, everyone has the same age", and B is the statement S(k+1): "in any group of k+1 people, everyone has the same age."

Step 4 says that all we have to do to complete the proof is assume A is true and try to derive B from it. Therefore, it is legitimate to use A and treat it as true in the process of proving B.

That's exactly what we're doing here. At this stage in the proof, we have two things are disposal:
we are assuming that, in every group of k people, everyone has the same age, and
we now have a particular group G of k+1 people, and are trying to use the above assumption to prove that everyone in G has the same age.

Within G, there happens to be a group of k people (namely, the group of everyone in G except for the one person P).

Since we are operating under the assumption that, in every group of k people, everyone has the same age, it follows that everyone in this smaller group within G has the same age.

And, it is perfectly legitimate to use this knowledge about this smaller group within G to try to show something about G itself, which is what the subsequent steps in the proof attempt to do.


Sakser fra http://www.math.toronto.edu/mathnet/falseProofs/guess25.htm

En annen festelig side for matte og logikk feil er :
http://members.cox.net/mathmistakes/

[knuta]

Ok, greit, men det går uansett i dass for k=1.

[Mr_Pin]

Stemmer bra det. Vi har bevist at alle er like gamle i en grupe med en person i, men aldri for en gruppe med to personer og kan dermed ikke vise at det stemmer for grupper større enn to heller. Eller som fasit sier :

Congratulations; you have correctly identified the fallacious step!
There might not be anyone else in G other than P or Q!

Remember that G is a group of k+1 people. As long as k > 1, k+1 > 2 and in this case there does exist a third person R in G. And, the rest of the proof will work.

So what does this prove? It proves that S(k) implies S(k+1), as long as k>1. In other words, it proves that if S(2) is true, then so is S(3), and if S(3) is true, so is S(4), and so on.

But nowhere have we shown that S(2) is true! The basis step of this induction showed that S(1) was true, but our induction step does not show that S(1) implies S(2).

Thinking about it in the following way may help. What this proof boils down to is this. It is showing that if every pair of people has the same age, then so does every group of three people, and so does every group of four people, and so on. But that's not anything startling; it should be fairly obvious if you think about it.

The problem is that it is not true that every pair of people has the same age! It is true that, in any group consisting of just one person, everyone in the group has the same age. However, the proof breaks down when you try to use that fact to prove that every pair has the same age. If {P,Q} is a pair of people, there's no third person R in the pair to make this step of the argument work, and therefore no way to conclude that P and Q have the same age by comparing them each to R, which is what the rest of the proof tries to do.

So: S(1) is true, and S(2) implies S(3) which implies S(4) etc., but S(1) does not imply S(2) and that is why the proof is fallacious.

Helt enig i at den første er best, men jeg klarte ikke å grave opp noe bedre på kort varsel.