# Reduce the Boolean expression (a+b+d+j) * 2.152?

Discussion in 'Homework Help' started by rssatnam, Feb 16, 2014.

Feb 16, 2014
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Reduce the Boolean expression (a+b+d+j) * 2.152 ?

2. ### WBahn Moderator

Mar 31, 2012
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What does it mean to multiply a Boolean expression by 2.152?

What is it you are looking for? Someone to do your work for you?

Feb 16, 2014
4
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I'm studying logic gates, and I'm able to solve some of the expressions but not this one, and I've to submit my assignment tomorrow so please if you can help me.

Thanks

4. ### WBahn Moderator

Mar 31, 2012
18,085
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Fine.

(a* 2.152)+(b* 2.152)+(d* 2.152)+(j* 2.152)

Until you give some idea of what it means to multiply a Boolean expression by a number, that's the best I can do.

5. ### Austin Clark Active Member

Dec 28, 2011
409
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I imagine you'd treat 2.152 as a constant TRUE or HIGH value.

6. ### WBahn Moderator

Mar 31, 2012
18,085
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Why? On what basis? Would we treat 0.01 as a constant TRUE? How about 0.95? Or 1.05? Or 94.3?

Without any context, it is meaningless.

And the OP obviously doesn't have any interest in providing a meaningful context for the question they asked. Until they do, they don't deserve (nor can they hope to get) a meaningful answer.

7. ### Austin Clark Active Member

Dec 28, 2011
409
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anything over 0 is considered a 1. Negative numbers would be a little trickier...
If we assume anything over 0 is a 1, and 0's remain 0s, then boolean algebra actually works out.

8. ### tshuck Well-Known Member

Oct 18, 2012
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That it's a very big assumption. Boolean algebra has no meaning for something like 2.152, unless that is being mapped into true, or false, as one of the two possible states, and that needs to be explicitly stated before any real meaning can be made of it.

9. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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Ok. IMHO this is not a Boolean expression. So much of the discussion is moot.

10. ### Austin Clark Active Member

Dec 28, 2011
409
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That's the thing, the numbers ARE being mapped to TRUE or FALSE. If the number is >0, it's TRUE, and if the number = 0, it's FALSE.

If you multiply anything by 0, you get 0, and if you multiply any two positive numbers, you get a positive number. Just like the AND function.

If you are adding only positive numbers and 0's, if any of those numbers are positive, the output is positive. Just like the OR function.

This is why we use the symbols we use for AND and OR functions. This is the thought process and mathematical model/reasoning George Boole used to develop boolean algebra in the first place.

11. ### WBahn Moderator

Mar 31, 2012
18,085
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But you are invoking a convention that is outside of Boolean algebra and it is not universal. And notice the additional caveats you had to put on it, namely that if we are adding only positive numbers. Well, what prevents us from adding a positive and a negative number? In addition, this is not even an integer, but a fractional number. And keep in mind that not all floating point representations are even capable of exactly representing zero to begin with.

12. ### Austin Clark Active Member

Dec 28, 2011
409
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You simply can't use negative numbers, as far as I can tell.

The numbers don't have to be integers for this to work.

This has nothing to to with floating point numbers, or representation of numbers within' a computer at all. This is simply mathematical theory.

13. ### WBahn Moderator

Mar 31, 2012
18,085
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If it has nothing to do with the representation of numbers to work, and if they don't even have to be integers for it to work, then why won't it work with negative numbers?

If this is just simple mathematical theory, then please give a reference to the theory that supports it.

And if we can use non-negative fractional values in Boolean expressions, then you better be prepared to explain how that works when you get to properties that Boolean algebra possesses that normal algebra doesn't. In particular, the fact that OR distributes over AND while addition does NOT distribute over multiplication.

14. ### Austin Clark Active Member

Dec 28, 2011
409
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a(b+c) = ab + ac in boolean algebra.

Write out the truth table for this expression. You'll find that if you replace HIGH/1 with any positive number, and you do the math with ordinary algebra, and find the solution, every time the output should be 0, it will be 0, and any time the output should be HIGH/1, the answer will be a positive number.

Basically, it DOES work.

I'm still unsure if you could find a way to get negative numbers to work though. That, and I'm not sure if there's a function, using regular algebra, that will perform the boolean inverse function.

It's just an observation I made early on, and I suspect the idea went into the forming of boolean algebra to some extent.

15. ### WBahn Moderator

Mar 31, 2012
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So please show how it works with XOR?

If it doesn't, then the whole notion is bushwah.

16. ### Austin Clark Active Member

Dec 28, 2011
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The XOR function is the same as:
A*!B + !A*B
or
(A+B)*!(A*B)

In both cases my idea checks out, however, the only problem is that I haven't found a way (if there is one) to perform the inverse function via algebraic means. That is, is there an algebraic function that maps 0 to any positive number (doesn't matter which one), and maps any positive number to 0. Perhaps I could find something that works if I map negative values to 0? Hmmm... I'm looking at the possibilities now.

17. ### WBahn Moderator

Mar 31, 2012
18,085
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But what is !B.

Your whole premise seems to be that we can mix and match non-negative values and Boolean logic values as we wish and just reduce things at the end.

So if I have A=3.4 and B=2.1, you are saying this is okay because we can multiply 3.4 by 2.1 and THEN convert them to True and False. Okay, so what is A+!B using the values before converting to True or False. If we have to convert before combining, then we could define ANY set of values (or even symbols) to be True and some other set to be False and be right in the same position.

18. ### Austin Clark Active Member

Dec 28, 2011
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It was just a curiosity, really. Boolean algebra and ordinary algebra share certain characteristics. In fact, I'm looking and wondering if there's a way to perform boolean algebra entirely using ordinary algebraic rules. Basically, find algebraic functions that perform the boolean algebraic functions.

19. ### tshuck Well-Known Member

Oct 18, 2012
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Where was this stated in the OP? Why is it mapped that way? Who said?

You can't just come in and claim that this is what is going on without being told that this is the mapping.

Anything else is just an assumption.

20. ### Austin Clark Active Member

Dec 28, 2011
409
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Of course the OP didn't state that, the OP didn't know what he was doing to begin with.

It's mapped that way because it makes the most sense, and it works.
If you map the values in this way, ordinary algebra works with boolean algebra, they work out precisely the same way.
The only problem is with inversions and negative numbers. Perhaps there's a way to make those work as well. I should make another thread on this topic or something, could be interesting. We really need a mathematician to chime in here, but I don't think many roam these forums.