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#1 Help Me ! » Counting Question - Binomial theorem » 2008-03-29 06:44:14

shaoen01
Replies: 1

How do you convert a sum in closed form into using the summation symbol and using an ellipsis.

Qns 1:
n(1+x)^{n-1}

Thanks

#2 Help Me ! » Using Mathematical induction » 2008-02-27 02:30:24

shaoen01
Replies: 0

Hi all,

Previously, i asked what a tromino is and i worked some equations out. However, i am not sure if my proof is right. So if someone can help me take a look and point out to me my mistake would be fantastic.

Qns 1:
Use mathematical induction to prove that for any integer n more than and equals to 1, if one square is removed from a 2^n x 2^n checkerboard, the remaining squares can be covered by an L-Tromino.

My workings:
A tromino consists of 3 small squares.

So i let P(n) = (2^n x 2^n)-1 for n>=1--> Because the qns says 1 square is removed.

Basis step:
P(1)=(2x2)-1
      = 3 --> Which is true.

Inductive Hypothesis:
So i am now supposed to prove P(n+1)=

x
is true.

So by multiplication definition (i think):
n=dq for some integer q
And d (divisor)=3 since a L-tromino has 3 squares.
n=3q

is an integer because the product or multiplication of an integer is still an integer. And because the divisor is 3, the above workings proved that the remaining squares can be covered by a L-tromino.

#3 Re: Help Me ! » Proving using Mathematical Induction » 2008-02-24 03:34:47

shaoen01

Wow, thanks! cool

#4 Re: Help Me ! » Proving using Mathematical Induction » 2008-02-24 03:22:50

shaoen01

Thanks JaneFairfax for the explanation. So how do i complete the proof by verifying that the statement is true for n = 0, 1 and 2? Do i just substitute the values in to check?

#5 Re: Help Me ! » Proving using Mathematical Induction » 2008-02-24 02:35:04

shaoen01

What i meant is how did you get the equation "3^n + 3^n + 3^n"? Are you trying to say that "3^n + 3^(n-1) + 3^(n-2)" is actually equivalent to "3^n + 3^n + 3^n"?

#6 Re: Help Me ! » Proving using Mathematical Induction » 2008-02-24 01:47:13

shaoen01

I am sorry but how did you get this "≤ 3^n + 3^n + 3^n, which is 3^(n+1)"? I am a bit confused ...

#7 Re: Help Me ! » Proving the uniqueness of Binary Integer Representations » 2008-02-23 20:36:59

shaoen01

Hi JaneFairFax:

I am sorry but i didn't quite understood this line from your reply:

. I do not get what "n>=2^r"? I don't think i have touched the division algorithm yet, the only division theorem i know is n=dq + r --> Is that the same?

Ricky: I am sorry, i didn't manage to catch your explanation.

All: I have just started on basic discrete mathematics, so please be patient with me. Thanks! smile

#8 Help Me ! » Proving using Mathematical Induction » 2008-02-23 20:29:32

shaoen01
Replies: 8

Hi all,

I hope someone can help me out with these 2 questions i have encountered.

Qns 1:
Use mathematical induction to prove that for any integer n more than and equals to 1, if one square is removed from a 2^n x 2^n checkerboard, the remaining squares can be covered by an L-Tromino.

My qns:
What is a L-tromino? And could someone give me some clue on how to work this out?

Qns 2:
Suppose that

is a sequence as follows:
for
.

(a) Prove that

for all
.

(b) Suppose s is any real number satisfying

(This implies that s>1.83). Prove that
for all
.
What i have done for Qns 2(a):
I have proven that the base step is true for k=3.
If i am not wrong i think i must prove that
is true. But i do not know what to do after this step:
. After peeking at the solution, it continues from
--> I do understand how
is derived. But what i don't understand is how did the solution derive the relationship of
. I feel that i am unable to spot or make this kind of relationship if i am given a similar question. How does one spot it?

#9 Help Me ! » Proving the uniqueness of Binary Integer Representations » 2008-02-23 04:23:24

shaoen01
Replies: 5

Hi all,

I am having some problems understanding how to prove the uniqueness of Binary Integer Representations and i hope someone can help me try to understand it better.

Qns:
Given any positive integer n, n has a unique representation in the form of

where r is a nonnegative integer,

Given Solution:
I was told to suppose that there is an integer n with two different representations as a sum of nonnegative integer powers of 2. Equating the two representations and canceling all identical terms gives:


where r and s are nonnegative integers, r<s and each
and each
equal 0 or 1. But by the formula for the sum of a geometric sequence:


which contradicts "

". Hence, the supposition is false, so any integer n has only one representations as a sum of nonnegative integer powers of 2.
[/math]

What i don't understand:
What does this statement mean exactly "integer n with two different representations as a sum of nonnegative integer powers of 2"? How do we derive

? And why is there a contradiction? I do not understand how the inequality equation was formed.

#10 Re: Help Me ! » Number Theory » 2008-02-16 17:03:46

shaoen01

Hi,

But how do you know that it will only be a prime when "3n-1 or n-1 is equal to 1"? Is something to do with the definition of prime, where m>1 and m=(r)(s), where r=1 or s=1?

#11 Re: Help Me ! » Number Theory » 2008-02-16 05:47:03

shaoen01

Hi Ricky,

Do pardon me if i didn't manage to catch what you were saying correctly. I know a number cannot be prime if it has a factor like numbers 4, 6 have factor like 2 and 3. And prime numbers can only be divided by themselves. Is that what you are trying to say? I know if i input values like n=3 or n=5 into the equation and it would not be prime answer. So i guess this is how it is proven?

#12 Help Me ! » Number Theory » 2008-02-16 04:57:08

shaoen01
Replies: 5

Hi all,

I am having some difficulty understanding the question below and did not quite get one part of the solution. It would be great if someone could guide me.

Qns:
Find the truth set of "

is prime" where the domain is N (special symbol N).

What i did and don't understand:
I know for a number m to be a prime number, m>1. I know that

. How exactly do i find out the truth set? What i don't understand about the solution is that it states n>2 is not prime. Why? How do i deduce that other than trying to substitute values in myself?

Thanks

#13 Re: Help Me ! » Fallacies in reasoning for modus ponens (method of affirming) » 2008-02-10 03:24:10

shaoen01

Thanks for your reply, i just wanted to make sure i was right.

#14 Help Me ! » Fallacies in reasoning for modus ponens (method of affirming) » 2008-02-09 17:40:52

shaoen01
Replies: 2

Hi everyone,

I have an invalid argument here, but i have difficulty determining if it is a converse or inverse error. It's hard for me to try to differentiate it at first glance, is there any technique in doing this right? Anyway, here is the argument:

Any sum of two rational numbers is rational
The sum r+s is rational
Therefore, the numbers r and s are both rational

Ans: Invalid argument, converse error.

My View:
By reading the above, i know it is wrong. But i can't explain it. Can someone maybe explain to me how it is a converse error? From my knowledge, a converse error is when:

p--> q
therefore, q
therefore, p

If someone can point to me which is p and q will be very helpful, thanks.

#15 Re: Help Me ! » Proof by using multiple theorems » 2008-01-30 03:57:19

shaoen01

I think the conclusion should be both are knaves because at step 6, there is a contradiction as shown below which proves that A is lying. So if A lies, then B is definitely not a knight.

   1. a
   2. therefore b -- A tells the truth
   3. therefore (a^~b) v (~a^b) -- B tells the truth (**)
   4. therefore a^~b -- elimination
   5. therefore ~b -- specialization
   6. therefore c (contradiction)
   7. therefore ~a -- by contradiction
   8. ~b -- A lies

#16 Re: Help Me ! » How do we determine if statement is vacuously true? » 2008-01-30 00:21:15

shaoen01

mathsyperson: So basically what you are saying is that since P(x) is always false, there is no way for P(x) --> Q(x) to be false right. Hence, it is vacuously true?

Ricky: I think i kind of get what you mean by the saying absolutely nothing part. In the case of my example, since the negation of the statement ("There is at least one ball in the bowl which is not blue" and the precondition is that there must be at least a ball inside) is false, so i just assume that the statement is true because the precondition could not be met. Am i right?

Thank you both for your replies.

#17 Help Me ! » How do we determine if statement is vacuously true? » 2008-01-28 06:15:18

shaoen01
Replies: 4

Hi all,

I was reading through my textbook and did understand the explanation of what it means for a statement to be vacuously true. But what i do not understand is the formal mathematical definition of it. Below is the scenario:

Suppose you have 5 blue and red balls and you place 2 blue balls and 1 red ball into a bowl. Consider the statement "All balls in the bowl is blue", is this true or false? The negation of this statement is "There is at least one ball in the bowl which is not blue". This requires at least one ball to be in the bowl, but there is none. So the negation is false; thus, the statement is true.

If P(x) = "x is in the bowl", Q(x) = "x is blue" then "for all x in D, P(x) --> Q(x)" is vacuously true if P(x) is false for every x in D.

My Question:
So what i don't understand is why must P(x) be false for every x in D for it to be vacuously true? Could we consider P(x) to be false when there is no ball in the bowl? Is this how you interpret it?

#18 Re: Help Me ! » Proof by using multiple theorems » 2008-01-26 17:40:43

shaoen01

JaneFairfax: I have one more question, i do understand step 1-4, but what i want to find out is why at step 5 do you put "~a" and not just put "~b"?

The definition of a contradiction rule: If you can show that the supposition that statement p is false leads logically to a contradiction, then you can conclude that p is true.

How do i apply the contradiction rule to derive the answer at step 5? I feel a little bit confused by the definition and unsure of how to apply it properly. Hopefully you can give me some insight, thanks.

#19 Re: Help Me ! » Proof by using multiple theorems » 2008-01-26 02:53:59

shaoen01

JaneFairfax: Thanks for showing me how to look at it from a different perspective. smile

#20 Re: Help Me ! » Proof by using multiple theorems » 2008-01-26 02:10:14

shaoen01

Wow, thanks your explanation has really helped me. So basically, when we are doing this kind of question i should find ways of contradicting the answer in some sense right? So that i can make further assumptions like the part on specialisation. At step 1, it is also possible for me to start with b instead a right?

#21 Help Me ! » Proof by using multiple theorems » 2008-01-26 01:08:49

shaoen01
Replies: 8

Hi all,

I have the solution to this question, but i don't understand how it was derived. And i hope someone could further explain it to me. A lot of theorems have been used such as modus tollen, proof by contradiction, etc.

Qns:
Knights always tell the truth, knaves always lie.
Can we tell what A and B are, from what they say below?

A: B is a knight
B: A and I are of opposite type.

Let a = "A is a knight"
Let b = "B is a knight"

Solution:

   1. a
   2. therefore b -- A tells the truth
   3. therefore (a^~b) v (~a^b) -- B tells the truth (**)
   4. therefore a^~b -- elimination
   5. therefore ~b -- specialization
   6. therefore c (contradiction)
   7. therefore ~a -- by contradiction
   8. ~b -- A lies


The above argument tells us that if there is a solution, it must be that both A and B are knaves.

My question:
From (**) onwards, i start to not understand the solution. Firstly, how do we know which we should eliminate at 3? When specializing at 5, why can't we select "a" instead of "~b". Do i just select anyone? And how does it lead to "c" from "~b" at 5 to 6?


Thanks for your help in advance!

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