Discrete Math

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.

 

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.

 

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!

 

Thank you so much for your help

 

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.

 

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.

 

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.