Group Theory and Linear Algebra Thread

[QuantumJG]

Hey this is a thread for my Group Theory and Linear Algebra subject.

Anyway:

Prove that if s.t. & then

How I did the proof:

where

where





Now here is where I'm not sure if I can do this.

&

//


With the next question I don't know how to show uniqueness.

Show that the remainder & quotient in Theorem 1.2.1 are unique.

Theorem 1.2.1:

If and . There are unique integers , such that

where

[kamil9876]

yes, is correct. This has two solutions: (1,1) and (-1,-1). You have to rule out the latter. e.g: if a=1, b=-1 then it is true that a|b and b|a in Z (showing the significance of the restriction to positive integers)

To get you started on the next one:

Suppose that qm+r=q'm+r'

therefore

m(q-q')=r'-r

Now you are ready to show that r'-r=0 as u want. (Hint: try to bound r'-r )

[TrueTears]

Hey this is a thread for my Group Theory and Linear Algebra subject.

Anyway:

Prove that if s.t. & then

How I did the proof:

where

where





Now here is where I'm not sure if I can do this.

&

//


With the next question I don't know how to show uniqueness.

Show that the remainder & quotient in Theorem 1.2.1 are unique.

Theorem 1.2.1:

If and . There are unique integers , such that

where

Hi, there, you are correct for your first question.

The 2nd question is a classic, here is how I would attempt it:

The theorem (If a and b are integers with b> 0, then there is a unique pair of integers q and r such that a = qb + r and 0 <= r < b) has 2 parts.

One we must prove existence of a 'r' such that 0<=r<b

Second we must prove uniqueness.

Let us first prove existence.

First we prove existence. Let . This set of integers contains non-negative elements (take ), so is a non-empty subset of ; by the well-ordering principle has a least element, which has the form for some integer . Thus with . If then contains a non-negative element
; this contradicts the minimality of , so we must have .

Now to prove uniqueness

To prove uniqueness, suppose that with and , so . If then , so , which is impossible since and lie between and inclusive. Hence and so .

[kamil9876]

notice also that this theorem extends to ie: if we let be real but keep as an integer.

[QuantumJG]

I saw my lecturer about this question today, but I forgot how he proved it.

Let [Tex] a,b,c \in \mathbb{Z} with gcd(a,b) = 1. Prove that if a|c and b|c then ab|c.

[/0]

Well if you write and as a products of factors, since and share no common factors, you can multiply them together and they must still divide .

[TrueTears]

I saw my lecturer about this question today, but I forgot how he proved it.

Let with . Prove that if a|c and b|c then ab|c.

We have ax + by = 1, c = ae and c = bf for some integers x, y, e and f. Then c = cax + cby = (bf)ax + (ae)by = ab(fx + ey), so ab|c.

[QuantumJG]

I saw my lecturer about this question today, but I forgot how he proved it.

Let with . Prove that if a|c and b|c then ab|c.

We have ax + by = 1, c = ae and c = bf for some integers x, y, e and f. Then c = cax + cby = (bf)ax + (ae)by = ab(fx + ey), so ab|c.

Yeah that's how he proved it. Thanks!

[QuantumJG]

I'm having trouble with this:

Show that if  with (mod , then the equation



has no solutions if

WTF!

[brightsky]

Actually forget mod 4, do it in mod 8.

We know that given a whole number is odd, then we have .

Likewise (tweaked from mod 4 case), if is even, then we have two cases:

1. or

2.

From here, by brute force, we can work out that the possibilities for can be any number other than 7.

[kamil9876]

mod 8:

:0 1 2 3 4 5 6 7

:0 1 4 1 0 1 4 1

(notice how it is symmetrical since , saves you half the time)

so amongst you cannot have no 4 or just a single 4 since otherwise the sum would be too small. So you need at least two 4's. But two 4's is 0 so that also makes your sum too small.

[pooshwaltzer]

Please suggest a possible financial application for said theory. Thanks.

[QuantumJG]

mod 8:

:0 1 2 3 4 5 6 7

:0 1 4 1 0 1 4 1

(notice how it is symmetrical since , saves you half the time)

so amongst you cannot have no 4 or just a single 4 since otherwise the sum would be too small. So you need at least two 4's. But two 4's is 0 so that also makes your sum too small.

Thanks Kamil. There was a question earlier where we had to do a multiplication table for 8 and I noticed that symmetry.

[QuantumJG]

This is a problem from last years exam.

Let V = and T: V V be the linear transformation defined as:

T(p(x)) = p(x+1) + 3p'(x)

a) Find the matrix T with respect to the basis {} for V.

So I got:

p(1) = {1,0,0}
p(x) = {2,1,0}
p(x²) = {1,4,1}

So I got T to be:



I just want to make sure this is right!

[kamil9876]

you want to be finding . I have no idea what you mean by etc., typo?