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 Ask a Mathematical Physicist

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v1n0d
post Nov 27 2013, 02:02 PM

Another roof, another proof.
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Excellent thread, keep 'em coming!
v1n0d
post Nov 30 2013, 12:42 PM

Another roof, another proof.
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QUOTE(Critical_Fallacy @ Nov 30 2013, 02:53 AM)
Thanks Dr v1n0d! Please be a “Mathematician by day” in this thread so that you can provide some guidance to students if they need help, when I'm not around. notworthy.gif
*
Will try my best. blush.gif

Thesis submission in a week, so it's busy busy busy at the moment! sweat.gif
v1n0d
post Dec 10 2013, 11:18 AM

Another roof, another proof.
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QUOTE(RED-HAIR-SHANKS @ Dec 10 2013, 04:51 AM)
Not too long ago I've came across this intriguing proof that sparked my mind. There are two proofs that show 0=1. This is how it goes:
First proof:
Let x = 1
So x²= (1)²=1
By subtracting x from x², we get:
x²-x=0
x(x-1)=0
x=0,x=1
Ergo, 0=x=1

Second proof:
Let x=y
Multiply x on both sides:
x²=xy
Next, subtract y² from both sides:
x²-y²=xy-y²
By factoring out (x-y) on both sides and dividing them out:
(x+y)(x-y)=y(x-y)
x+y=y
Since x=y:
y+y=y
2y=y
Hence,
2=1
Subtracting 1 from both sides:
1=0

My question is, are both of the 2 proofs above plausible, even if there is something wrong with it, can somebody show me the correct way of dealing with it? icon_question.gif
*
The mistakes in both your examples boil down to one word - "or".
For the first one, x(x-1)=0 implies that x=0 OR x=1. Only one of these will hold. Since the derivation of the quadratic equation x^2-x=0 began with the assumption that x=, the result x=0 must be discarded.

As for the second example, (x+y)(x-y)=y(x-y) implies that the products on each side are equal. Hence x+y can take two possible values, namely x+y=y OR x+y=x-y. This leaves you with either x=0 or y=0 since 0 is the only value that fulfills the requirement that y=-y (this also satisfies the initial condition that x=y). Now look at your final equation, which yields 2y=y. You can only discard y if it has non-zero value, which it doesn't (as shown above).
v1n0d
post Dec 10 2013, 02:02 PM

Another roof, another proof.
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QUOTE(maximR @ Dec 10 2013, 01:21 PM)
» Click to show Spoiler - click again to hide... «


Thank you . I've been placing my hand in different cookie jars , not knowing which to delve deeper into . For now , I have the luxury to do so so I'll jump around different topics , and get some serious work done in January .

The problem which I posed is from the book titled ' Basic Concepts of Mathematics and Logic ' by Michael C . Gemignani , which covers logic , set theory , counting , numbers , functions , ordering , probabilities and Euclidean geometry . It's quite a compact book , not the best designed book but I find it quite entertaining ( although the content comprises mostly of paragraphs of words ) .

That problem comes after this :

In order to disprove the following statement , which statements would we actually prove ?

' Some triangles are isosceles . '

The answer is : All triangles are not isosceles . (the answer doesn't delve deeper into how to prove this statement)

After the topic on Mathematical Disproofs , I ventured into Conjunctions and Disjunctions , which I've seen when learning Mathematical Reasoning in SPM Mathematics , and Boolean Algebra in SPM Physics .

A question : If A , then B is equivalent to ~( A ∧ ~B ) , with the reasoning :

If A , then B , thus if A occurs B must occur as well , or it cannot happen that A occurs but B does not occur , it cannot happen that A  occurs and ~B occurs .

So this statement ~( ~A ∧ B ) is false , right ? Because if A does not occur , B can occur .
*
Proof need not be unique. This is something most students are unable to grasp. We're so used to duplicating results here in Malaysia, that we often ignore the beauty of alternative proof.
For the triangle question, the word "some" indicates that you need to show that not all triangles are isosceles. This of course is a monumental task, as you would need to test all triangles in existence to see if they are isosceles. To simplify the task, we prove the contrapositive (if A implies B, then not A implies not B). The solution given in your book is based on this approach. Try come up with the contrapositive statement for your question and we'll have a look at it. smile.gif
As for your second question, simple boolean computation will give you the desired result. However, if you wish to see the details, take the set-theoretic approach, which is to show that if you pick any element x in set A, then x must also be in set B.


QUOTE(maximR @ Dec 10 2013, 01:21 PM)
» Click to show Spoiler - click again to hide... «

At the Big Bad Wolf sale , I bought Calculus ( Pearson International Edition ) textbook for RM 10 , which is a huge college level textbook which covers a lot of concepts in Calculus . I'm still in the preliminaries section , I'm familiar with most of the topics in the preliminaries ( they are covered in SPM Maths / Add Maths ) so I'm learning Logic and Proofs . I've also printed out your tutorials , but not sure where to start with all the resources around me .  biggrin.gif

In the Calculus textbook , there are a few problems which I need help with ( college level textbooks only provide solutions for Odd-Number questions  sweat.gif  )

1 . We can show that a number divided by zero is meaningless . Suppose a ≠ 0 . If a/0 = b , then a = 0 , which is a contradiction . Now find a reason why 0/0 is also meaningless .

2 . Show that any rational number p/q , for which the prime factorisation of q consists entirely of 2s and 5s , has a terminating decimal expansion .

3 . Show that between two real numbers there is a rational number .

4 . (a) Use the Fundamental Theorem of  Arithmetic to show that the square of any natural number greater than 1 can be written as the product of primes in a unique way , except for the order of the factors , with each prime occurring an even number of times .

(b) Show that √2 is irrational . ( Hint : Try a proof by contradiction . Suppose that √2 = p/q , where p and q are natural ( necessarily different from 1 ) . Then 2 = p^2/q^2 , and 2q^2 = p^2 . Now use (a) to get the contradiction .

Your help is greatly appreciated .  notworthy.gif By the way , is this topic under Number Theory ? It's under preliminaries , the author didn't label this sub-topic .
*
Yes, these questions are under the field of number theory. The author may not have labelled it because number theory tends to overlap with many other fields.
As for the questions you posed, I have the complete solutions for all of them, but I'm afraid that they're in storage on campus. I do however recall that the proof for question 4 (a) is available online, and it merely involves quadratic expansion of the original equation for the FTA.
v1n0d
post Dec 13 2013, 08:23 AM

Another roof, another proof.
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QUOTE(delsoo @ Dec 12 2013, 11:26 PM)
The new stpm system which also known as stpm modular system lacks a lot of detail explanation.... Whereas the old stpm syllabus book contains a lot of detailed explanation....
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The new STPM syllabus is very badly organized. I taught STPM for close to 3 years, and in my opinion, you should always start with the section on number systems before progressing to the later chapters.
v1n0d
post Dec 13 2013, 02:14 PM

Another roof, another proof.
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QUOTE(Critical_Fallacy @ Dec 13 2013, 12:04 PM)
Conventionally, mathematicians will tell you that there is no number that you can multiply by 0 to get a non-zero number. Hence, there is NO solution, and so any non-zero number divided by 0 is undefined or rather meaningless.
*
I think the full work should look something like this?

Let a,b,c∈R.
Suppose that a≠0 and pick any multiple b of a.
Then ∃ c ∋ a/c=b.
Thus a=bc.
Assume that c=0.
Then a=b(0) i.e. a=0 (contradiction since we assumed that a≠0)
Hence a number cannot be divided by zero as it fails to comply with the laws of divisibility.

P.S. Did this in a rush, do correct any mistakes if you spot them. tongue.gif

This post has been edited by v1n0d: Dec 13 2013, 02:15 PM
v1n0d
post Dec 13 2013, 04:54 PM

Another roof, another proof.
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QUOTE(K-o-E @ Dec 13 2013, 04:05 PM)
holy smokes...the stuffs you guys wrote are like alien to me. Good work extending the knowledge. Perhaps if my kids need tutorial, i can point to this thread. Anybody tutoring secondary students?
*
Everybody has their own forte, as do you sir. biggrin.gif
v1n0d
post Dec 13 2013, 09:20 PM

Another roof, another proof.
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Does anyone know if LYN accepts TeX commands? It would be a lot easier to put forward stuff with the proper formatting.
v1n0d
post Dec 13 2013, 10:13 PM

Another roof, another proof.
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QUOTE(maximR @ Dec 13 2013, 09:54 PM)
Yes , an equation which combines i , pi , e , 1 and 0 !  smile.gif Very beautiful equation .

I would appreciate it if you could provide tutorials on Logic .

https://fbcdn-sphotos-h-a.akamaihd.net/hpho...650586348_n.jpg

The link will bring you to a page of 2013 Further Mathematics STPM Repeat Paper . In Q1 , it asks for the validity of propositions , but in my book , it says that a proposition and an argument are two different things . A proposition can be true or false , valid and invalid are incorrect terms to refer to propositions .

I know how to do (a) and (b) , but what about the conclusion ?
*
The conclusion is that for every x∈R, ∃ an inverse y∈R ∋x+y=0.

Edit: Corrected! Thanks maximR

This post has been edited by v1n0d: Dec 13 2013, 10:36 PM
v1n0d
post Dec 13 2013, 10:15 PM

Another roof, another proof.
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QUOTE(RED-HAIR-SHANKS @ Dec 13 2013, 09:55 PM)
Hmmm hmm.gif I kinda fathom the question, but I have no idea on how to show that it's rational or not. A very tricky one I'd say.... unsure.gif
*
The solution makes use of the Gelfond-Schneider theorem. smile.gif

QUOTE(maximR @ Dec 13 2013, 10:15 PM)
Wait , I think I see something .

Suppose that (√2)^(√2) is irrational .

Let (√2)^(√2) = a , where a is an irrational number .

Raising the powers of both sides by √2 , we have (√2)^2 = a^(√2) ;

a^(√2) = 2  ( a is irrational , √2 is also irrational but 2 is rational ) .

shocking.gif

Call me a nerd , but this is much more exciting than a lot of things I've seen !
*
Correct, but what happens when a≠b?

This post has been edited by v1n0d: Dec 13 2013, 10:16 PM
v1n0d
post Dec 13 2013, 10:24 PM

Another roof, another proof.
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QUOTE(maximR @ Dec 13 2013, 10:18 PM)
Didn't think of that .  sweat.gif

Is there a general proof ?
*
It's not necessary to provide general proof. Existence is shown if you can provide just one example where the property holds.

QUOTE(maximR @ Dec 13 2013, 10:23 PM)
∋ means 'such that' ?

Any reasoning behind this , maybe a more detailed explanation ? I'm sorry , I've a lot to go .
*
Sorry, creature of habit. ∋ means such that. For any real number, an additive inverse exists within the set so that that number added to it's inverse will yield the additive identity, 0. This is a basic property of the vector space R of real numbers. You can find more info on this in books on linear algebra.

This post has been edited by v1n0d: Dec 13 2013, 10:26 PM
v1n0d
post Dec 13 2013, 10:35 PM

Another roof, another proof.
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QUOTE(maximR @ Dec 13 2013, 10:32 PM)
Can I say :

For every x∈R, ∃ an inverse y∈R such that x+y = 0 instead of x+y∈R ?
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Yes. In fact, you've just pointed out a mistake I made in my original post. I'll go correct it now. doh.gif

QUOTE(RED-HAIR-SHANKS @ Dec 13 2013, 10:33 PM)
Wait, did I miss something? shocking.gif  If I were to multiply an irrational number(√2) with itself, then I'll get a rational number, which is 2. So, the product of two irrational numbers is rational. Did I left out anything vital? unsure.gif
*
Your concern for the question is this - is the irrational power of an irrational still irrational?
v1n0d
post Dec 13 2013, 10:56 PM

Another roof, another proof.
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QUOTE(RED-HAIR-SHANKS @ Dec 13 2013, 10:52 PM)
An irrational to an irrational power may be rational, as what has been shown previously by maximR,(which also involves multiplying the Gelfond–Schneider constant with √2)

But, if in the case of √2*√3, I don't think the result is rational. Correct me if I'm wrong though.
*
The product of irrationals aren't always rational. As with the original question, one only need show an example to prove this statement true.
v1n0d
post Dec 13 2013, 11:01 PM

Another roof, another proof.
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QUOTE(maximR @ Dec 13 2013, 10:59 PM)
For every x and y such that both are irrational , their product is a rational number .

If the statement is revised as above , can one example prove the statement ?
*
You mean disprove. Yes, by providing a counterexample.
v1n0d
post Dec 13 2013, 11:07 PM

Another roof, another proof.
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QUOTE(RED-HAIR-SHANKS @ Dec 13 2013, 11:04 PM)
Oh, thanks for your clarification.
But, again, does this proof is constructive since √2^√3, and that the results is irrational?
*
I don't quite understand this bit. Mind rephrasing the question?
v1n0d
post Dec 14 2013, 08:49 AM

Another roof, another proof.
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QUOTE(RED-HAIR-SHANKS @ Dec 13 2013, 11:18 PM)
I'm sorry, my bad. What I want to point out was that, do we only need to show just one example/proof to imply that the statement is true or false?
Like in the case of a^b, and that both a and b is irrational(suppose that both a and b is √2). And that if we multiply (a^b)^√2, we will get 2, which is rational.
But on the other hand, if a=√2, and that b=√3, the results will surely be irrational. Now, doesn't both examples contradict each other? So, how does one tell or pick one of these examples that will proof that the statement is true? unsure.gif
*
Sorry, I slept off last night. sweat.gif
I see that Critical_Fallacy has already answered your question. laugh.gif
v1n0d
post Dec 15 2013, 09:32 PM

Another roof, another proof.
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QUOTE(maximR @ Dec 15 2013, 09:27 PM)
» Click to show Spoiler - click again to hide... «


Done with SPM . Starting on more 'advanced' concepts for now to get used to them before PreU . How are the lecturers at  USM by the way ? How 'rigorous' is the curriculum for Mech Eng ?
*
USM has brought in plenty of good lecturers due to their apex status. If you're gunning for Mech Eng, I'll suggest UTM as that's their premier course. Most of the other public unis in Malaysia have adopted their syllabus and local reference material was written by lecturers there. They also have strong ties with the industry. If I remember correctly, they signed an MoU so that their Pure Mech degree is recognized by MIT.

This post has been edited by v1n0d: Dec 15 2013, 09:33 PM
v1n0d
post Dec 15 2013, 09:39 PM

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QUOTE(maximR @ Dec 15 2013, 09:35 PM)
I'm not into Mechanical Engineering , I'm more geared towards Physics at the moment . This might change , that's why I'm keeping my options open , planning to do A Levels first ( which means IPTA would be out of the picture for me ) , and I've to really dig deep for scholarships after A Levels .
*
When it comes to Physics, my heart is with UM. I spent a couple of days there in the IPO selection camp, it still ranks as one of my top 3 academic experiences.
v1n0d
post Dec 15 2013, 10:10 PM

Another roof, another proof.
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QUOTE(maximR @ Dec 15 2013, 10:03 PM)
Do you know how I can be involved with the camp ? I've heard of the IMO one but information about IPO is scarce . sad.gif

How do you think I should prepare for it ? IPO involves a lot of college level Physics , not sure if I can manage to self-study everything .
*
Honestly, I have no idea. My F6 physics teacher picked my buddy and I to take the statewide exam and we scored. Next thing you know, we were part of 40-student pool for national team selections.

Most of the material covered in the selection camp was college level, so we were pretty lost. I do remember an experiment were asked to approximate the value of g without using a stopwatch.
v1n0d
post Dec 15 2013, 10:17 PM

Another roof, another proof.
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QUOTE(maximR @ Dec 15 2013, 10:15 PM)
I see . Where are you studying right now , if you don't mind me asking .
*
UTM. biggrin.gif

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