Discrete Maths Last Minute questions (1 Viewer)

Drsoccerball

Well-Known Member
Joined
May 28, 2014
Messages
3,657
Gender
Undisclosed
HSC
2015
Thanks mainly Integrand for saving me from dropping out of my degree by helping me in maths 1B but I need everyone's help just a bit longer for discrete maths.











 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Thanks mainly Integrand for saving me from dropping out of my degree by helping me in maths 1B but I need everyone's help just a bit longer for discrete maths.











3), Yeah, prime factorisation is unique. So every positive integer has only one prime factorisation. So if n_1 and n_2 have different prime factorisations, they cannot be the same number.
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007




lel LaTeX isn't working.
Claim:
A x (B u C) = (A x B) u (A x C)

If (x,y) is in the LHS then by definition x is in A and y is in either B or C.
If y is in B, then by definition (x,y) is in AxB. Similarly if y is in C then (x,y) is in AxC.
Hence LHS c RHS.

If (x,y) is in the RHS then either x is in A and y is in B, or x is in A and y is in C.
In either of these cases, x is in A and y is in BuC. Hence RHS c Ax(BuC) = LHS.

Hence LHS=RHS.
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Suppose a and b are congruent mod m, then we can write a = b + km for some integer k. Now, let d1 = gcd(a,m), and let d2 = gcd(b,m). Since d2 | b and d2 | m, we have d2 | (b + km), so d2 | a. So d2 divides both a and m, so d2 ≤ d1. A similar argument shows d1 ≤ d2 (since b = a + Km, where K = -k). So d1 = d2, as required.
 
Last edited:

Drsoccerball

Well-Known Member
Joined
May 28, 2014
Messages
3,657
Gender
Undisclosed
HSC
2015
Don't worry about the first question but the second is messed up

nvm got it.
 
Last edited:

Drsoccerball

Well-Known Member
Joined
May 28, 2014
Messages
3,657
Gender
Undisclosed
HSC
2015
A security agency sends messages using 12 letter code words, made up using the 26 letters of the English alphabet.

How many words contain at least one X and at least one Y and at least one Z?

I had :



Answer:
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
A security agency sends messages using 12 letter code words, made up using the 26 letters of the English alphabet.

How many words contain at least one X and at least one Y and at least one Z?

I had :



Answer:
What was the reasoning for your answer? You can use the inclusion/exclusion principle to get the answer.
 

Drsoccerball

Well-Known Member
Joined
May 28, 2014
Messages
3,657
Gender
Undisclosed
HSC
2015
What was the reasoning for your answer? You can use the inclusion/exclusion principle to get the answer.
That's what I did.

Unrestricted - (If 1 of the letters is in it) + (if pairs of the letters are in it)
 

Drsoccerball

Well-Known Member
Joined
May 28, 2014
Messages
3,657
Gender
Undisclosed
HSC
2015
Also I am doing past papers and circling questions I cant do Ill post them here after I finish all of em
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
^Why is it that virtually all of them are from the past paper book lol


But in regards to Q2, I also have that problem.

 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
How is this a necessary condition?
What necessary condition? Q3 is just asking for an example.

An example would be f:R->R, f(x) = sin(x), with A = [0, 2*pi] and B = [2*pi, 4*pi].

Then f(A) = [-1,1] = f(B), but A∩B = {2*pi}. So f(A∩B) = {sin(2*pi)} = {0}, whereas f(A)∩f(B) = [-1,1].
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top