# Given the Boolean Variables x and y, what are the associated four literals?

Discussion in 'Homework Help' started by mathnewb87, Dec 12, 2015.

1. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
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i am not sure if i know what a literal is and a minterm, from what i read i think a literal in a table are the variables xy, xz, yx, zx, these are just examples and the minterms are the 0's and 1's that correspond with them. Am i correct? i am answering questions about the boolean algebra form

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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if x and y are variables that can take one of the two states, 0 or 1, then the four states are:
xy
00
01
10
11

I have no effing clue what literals are.

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3. ### WBahn Moderator

Mar 31, 2012
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4. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
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so i had it backwards the literals are the true and false thank you this will help me. I am so lost to this stuff but i am going for video game design and apparently i need to learn this stuff. It is frustrating because I understand the meanings of a lot of the things i am learning, but when i have to use it in problems that is where may mind goes blank

5. ### Papabravo Expert

Feb 24, 2006
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You'll come across this terminology again and again. Just to cement the concept. If we are talking about integers, then 1377 is an integer literal. If we are talking about floating point numbers then 3.385e-12 is a very small floating point literal. If we are talking about a character, then 'a', '\$', and '+' are character literals. If we are talking about strings, then "Hello World" is a string literal. When we use a name like x, or y, or pay_rate to stand for something, we always have to ask what kind of something is it. As you learn about computer language and software the types of somethings you encounter will come from a fairly small set most of which I have already mentioned.

6. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
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so if there is something like this x y the 0's and 1's are the literals correct?
1 0
1 0
0 1
0 1

7. ### Papabravo Expert

Feb 24, 2006
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The four possible literals represented by a concatenation of x and y would be

00
01
10
11

That's all there are, and there aren't any more.

Last edited: Dec 12, 2015
8. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
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oh that is right the way i did it would be the negation correct?

9. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
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So for the midterms of xy would it be xy, xy', yx, yx'?

10. ### Papabravo Expert

Feb 24, 2006
10,338
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Not quite, you only have two distinct two character literals, namely 10, and 01.
The minterms are x'y', x'y, xy', and xy. Think all possible combinations, not permutations, of x and y.
Because the AND operator is commutative xy is always equal to yx.

There is no such thing as a midterm in Boolean logic; those only happen in freshman (1st year) classes.

11. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
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i meant minterm sorry about that stupid autocorrect i have a question on my boolean project about minterms and about sum-of-products, now from what i understand sum-of-products are all the combinations you can have with x,y and the addition and multiplication. Is that correct?

12. ### Papabravo Expert

Feb 24, 2006
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Again -- not quite.
The idea is that with two Boolean variables you can define sixteen different functions of two variables. Each of these sixteen functions can be described in either sum-of-products form or a product-of-sums form.
For example the exclusive-or function is true if x=1, or y=1, but is equal to 0 if x is equal to y. Let's write out the truth table

x y | xor
0 0 | 0
0 1 | 1
1 0 | 1
1 1 | 0

Looking at the truth table, I know that the minterms are defined by the rows in the table for which the value of the function is a 1. The exclusive-or function is described by the minterms x'y and xy', and if we OR those two minterms together we have a way to evaluate the exclusive-or function.

x ⊕ y = x'y + xy'

I also know that the maxterms correspond to the rows in the truth table for which the function is a 0. So the maxterms are (x' + y'), and (x + y). So the alternate definition of the exclusive-or function is:

x ⊕ y = (x' + y')⋅(x + y)

Now if two things are equal to the same thing then they must be equal to each other, so

x ⊕ y = (x' + y')⋅(x + y) = x'y + xy'

This very general result is true of all Boolean functions. They can all be decomposed into a SOP or a POS formulation and the formulations are equivalent.

Last edited: Dec 12, 2015
13. ### hp1729 Well-Known Member

Nov 23, 2015
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Boolean variables???
Like X, X-not, Y and Y-not

14. ### Papabravo Expert

Feb 24, 2006
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Literals are the values of Boolean variables.

15. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
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you lost me for a bit there but i think i am starting to understand the question is asking me
1. What is a sum-of-products? Given the Boolean Variables x and y, give an example of a Boolean function in sum-of-products form.
so what you did in the previous post was expand and simplify?

Mar 31, 2012
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Feb 19, 2010
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18. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
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i did read the link and the textbook, the textbook doesn't make sense, i am new to all of this so a lot of the textbooks i try to read are to complicated for me right now, the link helped but i like to make sure and usually when i ask people, their explanation is more simplified for me to understand, shteii01 if you aren't going to help don't respond with ignorant posts please thank you

19. ### mathnewb87 Thread Starter New Member

Dec 12, 2015
14
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thank you by the way, WBahn the link did help i apologize for not thanking you earlier

20. ### Papabravo Expert

Feb 24, 2006
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What I tried to show was that EVERY Boolean function can be represented as a truth table which enumerates all combinations of the variables and assigns a value to each combination. From that truth table, I can by inspection, write the sum-of-products form of the Boolean function. In addition I can also write the product-of-sums form of the function. Furthermore those two representations are equivalent forms that define the same Boolean function. I can also transform one representation into the other with only algebraic operations.