Discrete Maths Last Minute questions (2 Viewers)

leehuan · Nov 12, 2016

He-Mann said:
$x^2 + y^2 = 3k \implies 3 \mathrel{|}x^2, y^2 }$

Not convinced. Why?

Drsoccerball · Nov 12, 2016

leehuan said:
Not convinced. Why?

If you prove x^2 + y^2 = 9M for some M integer then it's proved but I don't know how to prove this that's why im resorting to exhaustion as the answers say but I don't understand it.

He-Mann · Nov 12, 2016

leehuan said:
Not convinced. Why?

some integer = k = (x^2+y^2)/3 = (x^2)/3 + (y^2)/3

Actually, I don't know. I'm going to puke, sorry. :'(

leehuan · Nov 12, 2016

Drsoccerball said:
If you prove x^2 + y^2 = 9M for some M integer then it's proved but I don't know how to prove this that's why im resorting to exhaustion as the answers say but I don't understand it.

Yeah I've seen that question before as well. I'm reluctant to give into exhaustion but if it appears in the exam I'm gonna do it because I have no choice

Drsoccerball · Nov 12, 2016

leehuan said:
Yeah I've seen that question before as well. I'm reluctant to give into exhaustion but if it appears in the exam I'm gonna do it because I have no choice

I showed my tutor my way and he said its all g but I have to prove something in regards to 3 mod 4.

Drsoccerball · Nov 12, 2016

He-Mann said:
$\begin{align*} \text{Question 1:} &\quad 1039x + 2013y = 4 \iff 1039x = 4-2013y \iff 1039x \equiv 4 \mod 2013 \iff x = 1039^{-1} 4 \mod 2013 \\ \text{Question 4}: & \quad 3 \mathrel{|} (x^2 + y^2) \implies \exists k \in \mathbb Z \quad s.t. \quad x^2 + y^2 = 3k \implies 3 \mathrel{|}x^2, y^2 \implies 3 \mathrel{|}x,y \end{align*}$

This isn't the general solution.

He-Mann · Nov 12, 2016

Drsoccerball said:
This isn't the general solution.

Well, find inverse of 1009 in mod 2013 and do the same thing for y. Alternatively, use Euclidean Algo to find GCD(1039,2013) then reverse process to get in the form GCD(1039,2013) = 1039a + 2013b

InteGrand · Nov 13, 2016

For the 3 | (x^2 + y^2), just look at all possible squares modulo 3: they are 0^2 = 0, 1^2 = 1, 2^2 = 1.

The only way to get two numbers from {0,1,1} to add to 0 (mod 3) is if both are 0. In other words, if two squares add to a multiple of 3, they are each 0 mod 3, as required.

Drsoccerball · Nov 13, 2016

Give an example of a relation on the set
{1, 2} which is both symmetric and antisymmetric. (It is sufficient to draw an arrow diagram to represent the relation.)

InteGrand · Nov 13, 2016

Drsoccerball said:
Give an example of a relation on the set
{1, 2} which is both symmetric and antisymmetric. (It is sufficient to draw an arrow diagram to represent the relation.)

Here's one (essentially a trivial one): just draw 1 and 2 (represented by dots for example) with no arrows whatsoever.

InteGrand · Nov 13, 2016

For less trivial ones, they'll be precisely those 'arrow diagrams' where only loops from a dot to itself exist. So there are four possibilities. E.g. you could do one with a loop from 1 to itself and a loop from 2 to itself (and no other arrows).

Drsoccerball · Nov 13, 2016

What is the negation of : $S_n \subset S_2 \cup S_3 ... \cup S_m$

Drsoccerball · Nov 13, 2016

ind the probability that a hand of 13 cards dealt from a standard pack
will have exactly one card in at least one of the four suits.

What is this even asking?

InteGrand · Nov 13, 2016

Drsoccerball said:
What is the negation of : $S_n \subset S_2 \cup S_3 ... \cup S_m$

$\noindent $S_{n} \not\subset S_2 \cup S_3 \cup \cdots \cup S_m$$

Drsoccerball · Nov 13, 2016

InteGrand said:
$\noindent $S_{n} \not\subset S_2 \cup S_3 \cup \cdots \cup S_m$$

Definition of not cumulative :
$\forall \ m \ \geq 2 \ \exists \ n \ > m \ $\noindent $S_{n} \not\subset S_2 \cup S_3 \cup \cdots \cup S_m$$$

Prove that the sequence S_2,S_3...

where S_n = {multiples of n} is not a cumulative function.

I let n = m + 1 is this right ?

InteGrand · Nov 13, 2016

Drsoccerball said:
Definition of not cumulative :
$\forall \ m \ \geq 2 \ \exists \ n \ > m \ $\noindent $S_{n} \not\subset S_2 \cup S_3 \cup \cdots \cup S_m$$$

Prove that the sequence S_2,S_3...

where S_n = {multiples of n} is not a cumulative function.

I let n = m + 1 is this right ?

Can't just use m + 1, since could have a factors be included in the lower sets in general. Instead, use n = a prime number bigger than m. I assume you can see why this'll work (also there infinitely many primes so this is well-defined).

Drsoccerball · Nov 13, 2016

Determine the number of cycles of length 4 that Q_n contains.

Drsoccerball · Nov 13, 2016

Can someone check my logic?

$Let A, B, and C be sets. Prove that if $ A - B \subseteq C $ and $ A \cap B \subseteq C $ then $ A \subseteq C.$

$$ Let $ x \in A - B \\ \Leftrightarrow \ x \ \in \ A \cap B^c$

$\therefore $ Either $ x \in A $ \Leftrightarrow x \in C \therefore A \subseteq C \ $ (Which completes the proof if true)or $\\ x \in B^c \Leftrightarrow x \in C \therefore B^c \subseteq C -- (1)$

$$ Let $ a \in A \cap B \\ \therefore a \in A \Leftrightarrow a \in C \therefore A \subseteq C \ $ (Which completes the proof if it was the case) or $\\$

$a \in B \Leftrightarrow a \in C \ \therefore B \subseteq C \ -- (2)$

$$ From (1) and (2) $ \ B \cup B^c \subseteq C \cup C$

$This implies that C is the universal set and thus any set A is always a subset of the universal set.$

[] ?

InteGrand · Nov 13, 2016

Drsoccerball said:
Can someone check my logic?

$Let A, B, and C be sets. Prove that if $ A - B \subseteq C $ and $ A \cap B \subseteq C $ then $ A \subseteq C.$

$$ Let $ x \in A - B \\ \Leftrightarrow \ x \ \in \ A \cap B^c$

$\therefore $ Either $ x \in A $ \Leftrightarrow x \in C \therefore A \subseteq C \ $ (Which completes the proof if true)or $\\ x \in B^c \Leftrightarrow x \in C \therefore B^c \subseteq C -- (1)$

$$ Let $ a \in A \cap B \\ \therefore a \in A \Leftrightarrow a \in C \therefore A \subseteq C \ $ (Which completes the proof if it was the case) or $\\$

$a \in B \Leftrightarrow a \in C \ \therefore B \subseteq C \ -- (2)$

$$ From (1) and (2) $ \ B \cup B^c \subseteq C \cup C$

$This implies that C is the universal set and thus any set A is always a subset of the universal set.$

[] ?

How did you get C being a universal set?

Drsoccerball · Nov 13, 2016

InteGrand said:
How did you get C being a universal set?

$B \cup B^c \subseteq C$

$U \subseteq C$

$\therefore C = U$

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