Discrete Math

[Time To Buy]

I really need help with these problems

Number two or three is all I need help with. (Not both)

Number four and one of either five or six (Not both for 5 or six)


Thank you, it's really important to me.

 

[Alex_J_Leon]

Worked on the easiest one =]

5.a.

100 is positive so you simply convert 100 to binary...

You know how to do this right?
100/2 = 50 R 0
50/2 = 25 R 0
25/2 = 12 R 1
12/2 = 6 R 0
6/2 = 3 R 0
3/2 = 1 R 1
1/2 = 0 R 1

(You divide by 2 and record the result and the remainder, the binary conversion is produced going from the bottom to the top)

So your number is 1100100 but that is 7 bits so you add the leading zero.

Ans: 01100100

5.b.

-54 is negative so fancy things must be done!

First you convert the number to binary, same as before:
54/2 = 27 R 0
27/2 = 13 R 1
13/2 = 6 R 1
6/2 = 3 R 0
3/2 = 1 R 1
1/2 = 0 R 1

So POSITIVE 54 is 110110 in 6 bits. Add the leading zeroes to get 00110110

To get the answer to be -54, we must perform the two's complement:

Now you invert each of the individual bits.

00110110 -> 11001001

Then add one to the result

11001001 + 1 = 11001001

Ans: 11001001

5.c.

First, draw a conversion table for Hex to decimal that shows A=10, B=11, etc.

Then I'd simply write:

7*16^0 + 4*16^1 + 11*16^2 + 14*16^3

I don't know how to explain that, really... it's just how you do it =p.

 

[Alex_J_Leon]

I'm back... I did part of number 2.

2.a.

p implies q is only false when p = 1 and q = 0

Example: If I have $20, I will bang my gf TONIGHT

The premise, p is "if I have $20" and q, the conclusion, is that you would "bang your gf"

So for pq = 00... You are a broke wigga and you didn't get laid. Since you didn't get $20, you can only assume that p implies q was true... you have no way of proving otherwise.
For pq = 01... You are a broke wigga and you DID get laid. Again, you didn't get $20 so you cannot reject p implies q. Maybe you romanced her.
For pq = 10... You got $20 but you didn't get laid. In this situation, you broke the promise you made to yourself and instead bought weed with your money. You broke the promise, so p implies q is false
For pq = 11... You got the $20 and you got laid. Self-explanatory.

So the truth table is
p q p=>q
0 0 1
0 1 1
1 0 0
1 1 1

For NOT p OR q

p q ~p v q
0 0 1 (1 + 0 = 1)
0 1 1 (1 + 1 = 1)
1 0 0 (0 + 0 = 0)
1 1 1 (0 + 1 = 1)

Yep, equal.

2.b.

By Demorgan's Law (bubble logic):

NOT p OR NOT q = NOT(p AND q)

Then you invert it:

NOT(NOT(p and q)) the NOT gates cancel each other out and you are left with:
Ans: p AND q

2.c.

Cba to do this right now, I have to go!

 

[Time To Buy]

Thank you so much for your help

 

[win4fun]

I have to take this next year and these questions aren't making me that excited lol. How are you finding the course so far?

 

[Alex_J_Leon]

I have to take this next year and these questions aren't making me that excited lol. How are you finding the course so far?

It is actually a really trivial course, at least in my opinion. Back when I took it, I made straight 100's on the exams haha.

 

[Time To Buy]

I have to take this next year and these questions aren't making me that excited lol. How are you finding the course so far?

I'm not actually taking the course, which is why I needed the help. A friend of mine failed a test and this was his retake exam but he really needed help on it so he asked me. I could do some of them, but I just didn't know what the symbols in the other questions meant.

 

[Jei jei KK]

3 a)

reads "There exists a natural number m, so that for every natural number n, m is m + n".
This is false. Natural numbers are the set {1,2,3...} or {0,1,2,3...} depending on your definition. The only value of n with which this is true is 0 (m = m + n = m + 0 = m), and the statement explicitly claims that m = m + n for every natural number n. It's fairly obvious that m + n != m, with n = 1 or with any other non-zero n.

b)
The negation should be: "For every natural number m, there exists a natural number n, so that m != m + n". Or, written with quantifiers:

Because of the law of excluded middle, and having proven the statement false the negation must be true. Or just pick n = 1 or any other non-zero number. The claim holds with all of them.

c) The statement reads: "For all real numbers X, if x is less than or equal to zero, it follows that x + 1 is less than or equal to 1." The contraposition is: "For all real numbers X, if x + 1 is greater than 1, it follows that x is greater than 0".
This is true and rather self-evident.