Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

Fistfiz
2012-10-10 03:42:16

Anakin wrote:

Nevertheless, I did finish the question with all of your help and now it's onto this week's assignment! I'll be back if I get stuck. big_smile

I hope so (not that you get stuck... that you post more interesting questions big_smile ) since i'm studying the same subject right now. Good work!

Anakin
2012-10-09 15:11:51

Interesting. I suppose it is rather complete then, isn't it? At least considering the familiar example that you posted. smile

I don't recall reading it in a textbook either but I have seen that exact example being proven the way that you've proven it - albeit it can look a little confusing at first.

Nevertheless, I did finish the question with all of your help and now it's onto this week's assignment! I'll be back if I get stuck. big_smile

Fistfiz
2012-10-07 00:10:36

Yes, it's exactly the same argument but wrote in a less pedant manner big_smile
In my opinion, your demonstration in post#1 is not incomplete: i've read lots of (basic level) demonstration where "formal" things
are left to intuition and they're still considered good demonstration

For example consider the classical demonstration that sqrt2 is irrational:
we start with two proposition which are:

p: (a/b)^2=2
q: a and b are irreducible

and we prove that p AND q => NOT q;
but we have to assume q to be true, because in each case it's false we get back to a case in which is true.
so NOT q is false... so p AND q is false, but since q is true, p must be false.

But i've never (for God's sake) read a math's book explaining that in proposition in these terms.

Anakin
2012-10-06 15:10:44

Hi everyone,

Fistfiz: Yes, I think your formal logic work is correct. Actually, yes it is correct. My third demonstration, though correct, is a little incomplete. However, your post completes it. Luckily, I ended up using the same approach as you by using the 3rd demonstration with my second demonstration to make a complete proof. I'll post my solution below:

AA = A
det(AA) = det(A)
det(A) * det(A) - det(A) = 0
det(A) * [det(A) - 1] = 0

This implies that det(A) = 0 or det(A) = 1. We can accept the first case. As for the second case, it implies that A is invertible.
AA = A
A^(-1) AA = A^(-1) A
A = I

But one of our preconditions was that A != I, therefore we can ignore the second case because det(A) = 1 and AA = A <==> A = I. This concludes that if A != I, then det(A) must equal 0.

This is equivalent to what you wrote I think, just differently worded and without the use of formal logic.

Fistfiz
2012-10-06 05:19:42

ok thanks...

seems to me your (Anakin) third demonstration is right

call
p: AA=A;
p': A!=I;
q: detA=0
q': detA=1

we have that p=>(q OR q')
and q'=>NOT p' (or, equivalent, p'=>NOT q')
so p AND p' => (q OR q') AND (NOT q') = q

Do you agree?

EDIT: to be correct i think i should say:
q' AND p => not p' which is equivalent to p' => (NOT q' or NOT p) => NOT q'
because we assume p is true

bobbym
2012-10-05 22:18:04

(!=)That is computerese for not equal.

Fistfiz
2012-10-05 22:16:28

sorry anakin
what is "A!" ?

bobbym
2012-10-05 21:41:40

Yes, I am sorry. Very obvious.

Anakin
2012-10-05 21:38:39

Hi Bobbym,

IA = A as one of the properties of the identity matrix is that IA = AI = A (for an n x n matrix; a little different for m x n).

bobbym
2012-10-05 21:14:40

Hi;

I meant

A^(-1) AA

you have that reducing down to A, sholdn't it be IA?

Anakin
2012-10-05 20:46:50

I just found out (Google!) that there is a term for such matrices, they are called idempotent matrices. I'm going to look a little more into them now and see if something useful appears.

Anakin
2012-10-05 20:38:22

Oops you edited your post again before I was able to read it.

I don't believe there is an error except that I'm assuming that A is invertible, as in A^(-1) A = I. Which I suppose is the wrong way of doing it. For instance if we want to prove that X => Y. We can't assume (not Y) and assuming a part of X (in this case that AA = A) and then say that since another part of X (the A = I part) is not satisfied , we have a contradiction. Thus Y must be true.

That would just be counter-productive - which is why I am still unsatisfied with that solution. The other way I can do this is using the contrapositive:
det(A) != 0 (which means that A is invertible) IMPLIES that AA != A OR A = I.

But I didn't really get very far with that either.

bobbym
2012-10-05 20:21:41

Did you see the error in line 2?

Anakin
2012-10-05 20:20:14

That's alright, Bobbym. I'll see if someone else has something to add.

But thank you very much - as always - for the aid!

bobbym
2012-10-05 20:03:44

Hi Anakin;

I was unable to do that. Maybe someone else can help more.

Then, AA = A
A^(-1) AA = A^(-1) A
A = I

But I can say this there is an error in your second line of your 3rd attempt.

Board footer

Powered by FluxBB