A Summer of Maths

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Well, the theorem is not just saying that they are rational.. since the leading coefficient is 1, they are integer factors of

210 = 2\times3\times5\times7

You'd invoke the rational root theorem if you had to "find the roots of this polynomial" -- your question was to check whether the solutions were the prime numbers below 10, which is a simple verification proof (since we can find the primes less than 10, even using elementary methods, very quickly).

To clarify: RRT says "if there are rational roots, they must be 2, 3, 5 or 7"; it doesn't say that they're the only possibilities for roots, so in any case you'd have to check directly. And since your question only asks about 2,3,5,7, you try them, and they work -- no need for RRT.

(edited 11 years ago)

Reply 581

TwoTwoOne

Original post by jack.hadamard

I tend to like pure more than applied. Anyway, I don't think I would be able to comprehend much of this book at that stage. :smile:

(I'm willing to push hard enough to learn something advanced only if I would enjoy the topic. :tongue:

)

On a side note, are you planning to do changes on your notes?
I would like to print Groups and Analysis for now (to use them as a guide/reference of the material covered).

Just thought I'd try to persuade you to the dark side. :smile:

Although you can get pretty pure with this (i.e. dynamical systems) via differential geometry

Not major ones I think, just typos and corrections. I can never resist a pretty diagram though, so I might add one here and there, but apart from that they are fairly complete.

(edited 11 years ago)

Reply 582

jack.hadamard

Original post by nuodai

.. is a simple verification proof (since we can find the primes less than 10, even using elementary methods, very quickly).

I see your point, I agree. I should have phrased it better, but next time. :tongue:

Reply 583

jack.hadamard

Original post by TwoTwoOne

Not major ones I think, just typos and corrections. I can never resist a pretty diagram though, so I might add one here and there, but apart from that they are fairly complete.

Well, adding/changing few pages afterwards is easy. Thanks for the confirmation! :smile:

Reply 584

jack.hadamard

In order to insert something fruitful of more interest to others, I will steal two examples. :colondollar:

Proposion: If

n^2

is even, then

n

is even.

Proof (1):

Suppose that

n

is even, then

n = 2k

for some integer

k

. Hence,

n^2 = (2k)^2 = 2(2k^2)

shows that

n^2

is even.

Proof (2):

Suppose, on the contrary, that

n

is odd, then

n = 2k + 1

for some integer

k

. Hence,

n^2 = (2k + 1)^2 = 2(2k^2 + 2k) + 1

shows that

n^2

is odd.

Proposition: Every positive real numbers is greater than or equal to one.

Proof:

Let

r

denote the smallest positive real number. Observe that exactly one of

r < 1

r = 1

and

r > 1

is true by the Trichotomy Law.

If

r < 1

, then

0 < r^2< r

is in contradiction with our assumption that

r

is the smallest positive real.
If

r > 1

, then

0 < \sqrt{r} < r

-- again, this leads to a contradiction.

Therefore,

r = 1

.

Any comments? :tongue:

Reply 585

nuodai

Original post by jack.hadamard

Any comments? :tongue:

I'm not sure if I should answer this. (In fact, I think I should back out of this thread, it's for people who are about to start their undergraduate, not for people who have just finished it!) But if I am allowed...

Spoiler

(edited 11 years ago)

Reply 586

Dog4444

Original post by nuodai

Spoiler

Just to verify, is it because he proved not Q implies not P what is equivalent to P implies Q?

Reply 587

nuodai

Original post by Dog4444

Just to verify, is it because he proved not Q implies not P what is equivalent to P implies Q?

Yup. The assertion "if n² is even then n is even" doesn't necessitate in any way that n² ever has to be even. The important thing is that if n² is even then n can't have been odd (because if n is odd then n² is odd [which]), and thereafter it doesn't matter, at least for the purposes of proving the statement, what happens when n is even.

For instance, as ridiculous as it sounds, it's true to say "if 2n is odd then n is odd", and it's also true to say "if 2n is odd then n is even".

(edited 11 years ago)

Reply 588

Dog4444

Some simple logic questions here:

Is it right to say (x-2)^-4=0 => x=4? I mean, it does imply that x=4, but x=-2 as well. So, considering it's not a iff statement, is it technically correct?

And, is it right to write (x-2)^-4=0 => x=4, -2? While being true, it doesn't say that it's the only way (x-2)^-4=0 is true.

Or the only correct way is to write (x-2)^-4=0 <=> x=4, -2?

If, A => B, does it mean that B is sufficient condition for A (not the other way around)? (what means (x-2)^-4=0 => x=4, (x-2)^-4=0 => x=4, -2 are both technically correct? )

(edited 11 years ago)

Reply 589

DamoclesAustria

Original post by jack.hadamard

I digged another one of these nice Groups questions. :biggrin:

Question:

Let

G

be a group with identity

e

, and

\phi : G \to G

a function such that

\phi(g_1)\phi(g_2)\phi(g_3) = \phi(h_1)\phi(h_2)\phi(h_3)

when

g_1 g_2 g_3 = e = h_1 h_2 h_3

.

Prove that there exists an element

a \in G

such that

\psi(x) = a\phi(x)

is a homomorphism (that is,

\psi(xy) = \psi(x) \psi(y)

for all

x,y \in G

).

Hint:

Spoiler

Wow, it took me a whopping two weeks to get around to writing a solution to your problem...again. At least I have a legit excuse: I was in Bulgaria participating in the IMC, where I got a Second Prize :smile:

Anyway, here it goes.

Spoiler

In return, I have another very nice group question:
Show that every group of even order contains an element of order two.

(edited 11 years ago)

Reply 590

Lord of the Flies

Original post by Dog4444

Not sure about this so feel free to correct me - perhaps I'll learn something!

1.

(x-2)^2=4\Rightarrow x=4

is false. The solution statement x = 4 implies that you have given all possible values of x, which is not true. Even if say, the question specified x positive, I would write (x-2)^2 = 4 => x = {-2, 4} but x positive so x = 4

2. There is only one way

(x-2)^2 = 4

is true, and it implies x = {-2, 4}

3.

(x-2)^2=4\Leftrightarrow x=\{-2,4\}

is correct. The iff statement means an implication works both ways. Since this is the case, the statement is correct. This is also why (1.) must be false. When you can write an iff relation, writing either one implication or the other is also correct - the double arrow is a shorthand to writing both.

A\Rightarrow B

means that A is a sufficient condition for B to be true.

(edited 11 years ago)

Reply 591

nuodai

Original post by Lord of the Flies

Not sure about this so feel free to correct me - perhaps I'll learn something!

1.

(x-2)^2=4\Rightarrow x=4

(x-2)^2 = 4

is true, and it implies x = {-2, 4}

3.

(x-2)^2=4\Leftrightarrow x=\{-2,4\}

A\Rightarrow B

means that A is a sufficient condition for B to be true.

Everywhere where you've written

x=\{-2,4\}

, you should have written

x \in \{ -2, 4 \}

; otherwise what you say is mostly right. If you want to use conjunctions and disjunctions (where

\vee

essentially means 'or' and

\wedge

means 'and') then it'd also be fine to write something like

(x=-2) \vee (x=4)

. So for instance

(x-2)^2-4 = 0 \Rightarrow (x=-2) \vee (x=4)

and

((x-2)^2- 4 = 0) \wedge (x \ge 0) \Rightarrow x=4

are both true.

Original post by Dog4444

Is it right to say (x-2)^-4=0 => x=4? I mean, it does imply that x=4, but x=-2 as well. So, considering it's not a iff statement, is it technically correct?

Be very careful with what you mean by 'imply'. It doesn't imply that

x=4

, but it does imply that

x

could be equal to 4; however it could be equal to -2. There are many equivalent ways of stating this. Here are a few:

(x-2)^2-4 = 0

implies

x=-2

x=4

x=-2

and

x=4

are solutions to

(x-2)^2-4 = 0

(x-2)^2-4 = 0

implies

x \in \{ -2, 4 \}

...etc.

(edited 11 years ago)

Reply 592

jj193

Original post by DamoclesAustria

Anyway, here it goes.

Spoiler

In return, I have another very nice group question:
Show that every group of even order contains an element of order two.

Show that every group of even order contains an element of order two.

lemma 1.
If there is an element g of even order then g^(x/2) has order 2

Is the following true?
If every element of G has odd order then |G| is odd
sketchy proof: every term in the inclusion exclusion on G\{e} is even.

In the context of this question, the contrapositive of the second gives an element of even order.

second prize in IMC, well done!

(edited 11 years ago)

Reply 593

Lord of the Flies

Original post by nuodai

Everywhere where you've written

x=\{-2,4\}

, you should have written

x \in \{ -2, 4 \}

; otherwise what you say is mostly right. If you want to use conjunctions and disjunctions (where

\vee

essentially means 'or' and

\wedge

means 'and') then it'd also be fine to write something like

(x=-2) \vee (x=4)

. So for instance

(x-2)^2-4 = 0 \Rightarrow (x=-2) \vee (x=4)

and

((x-2)^2- 4 = 0) \wedge (x \ge 0) \Rightarrow x=4

are both true.

Thanks for correcting - I'll bear the semiotics in mind! :smile:

Reply 594

nohomo

Let

S

be a set of three, not necessarily distinct, positive integers. Show that one can transform

S

into a set containing

0

by a finite number of applications of the following rule: Select two of the three integers, say

x

and

y

, where

x \le y

, and replace them with

2x

and

y-x

(edited 11 years ago)

Reply 595

nohomo

Prove that, for any integers

a,b,c

, there exists a positive integer

n

such that

\sqrt{n^3+an^2+bn+c}

is not an integer.

Reply 596

Dog4444

Original post by Lord of the Flies

...

Original post by nuodai

...

Ok, thanks. And what about crazy stuff like, x=3 => x^2=9 => x=+/- 3 ? If the logic is right, we get a contradiction.

Reply 597

Dirac Spinor

Original post by Dog4444

Ok, thanks. And what about crazy stuff like, x=3 => x^2=9 => x=+/- 3 ? If the logic is right, we get a contradiction.

(plus OR minus) is consistent with (plus)

Reply 598

Lord of the Flies

Original post by Dog4444

Ok, thanks. And what about crazy stuff like, x=3 => x^2=9 => x=+/- 3 ? If the logic is right, we get a contradiction.

x=\pm 3

means +3 OR -3 so there is no contradiction.

Edit: didn't see the above post! :tongue:

(edited 11 years ago)

Reply 599

desijut

I have done nothing in my summer holidays except PS3, **** about and painting (dad got me oly tickets so I'm paying him back)

After results, when **** gets real, I may look back into this stuff called "maths"