23
$\begingroup$

All the definitions I came across so far stated, that if a statement is true, then also its dual statement is true and this dual statement is obtained by changing + for ., 0 for 1 and vice versa.

However when I say 1+1, whose dual statement according to the above is 0.0, I get opposite results, that is:

1 + 1 = 1
0 . 0 = 0

How should I understand this duality principle ?

$\endgroup$

6 Answers 6

23
$\begingroup$

"$1 + 1 = 1$" is a statement (a boolean statement, in fact), and indeed, $1 + 1 = 1$ happens to be a true statement.

Likewise, the entire statement "$0 \cdot 0 = 0$" is a true statement, since $0 \cdot 0$ correctly evaluates to false: and this is exactly what "$0 \cdot 0 = 0$" asserts, so it is a correct (true) statement about the falsity of $0 \cdot 0$.

The duality principle ensures that "if we exchange every symbol by its dual in a formula, we get the dual result".

  • Everywhere we see 1, change to 0.
  • Everywhere we see 0, change to 1.
  • Similarly, + to $\cdot$, and $\cdot$ to +.

More examples:

(a) 0 . 1 = 0: is a true statement asserting that "false and true evaluates to false"

(b) 1 + 0 = 1: is the dual of (a): it is a true statement asserting that "true or false evaluates true."


(c) 1 . 1 = 1: it is a true statement asserting that "true and true evaluates to true".

(d) 0 + 0 = 0: (d) is the dual of (c): it is a true statement asserting, correctly, that "false or false evaluates to false".

$\endgroup$
7
  • $\begingroup$ Ok, so I think I am slowly starting to get it, but I still have difficulties with the last sentence. The dual symbols are 1 and 0 and + and ., right ? And the dual result means, that if the original result was 1, the dual result will be 0. Is this correct ? $\endgroup$
    – jcxz
    Jun 1, 2013 at 12:45
  • 1
    $\begingroup$ Yes, exactly. The "result" changes, but the truth value of the entire statement remains true. $\endgroup$
    – amWhy
    Jun 1, 2013 at 12:48
  • $\begingroup$ ok, thanks very much, I think it is clear now $\endgroup$
    – jcxz
    Jun 1, 2013 at 12:50
  • 1
    $\begingroup$ Good: I can understand the confusion! $\endgroup$
    – amWhy
    Jun 1, 2013 at 12:51
  • $\begingroup$ @amWhy: Very nicely formatted answer and answer! +1 $\endgroup$
    – Amzoti
    Jun 2, 2013 at 1:59
3
$\begingroup$

The statement is the full equation, including the = sign. 1+1 is neither true nor false: it takes the value 1, but it is not actually saying anything. Analogously, the expression "Tom has a cat" is neither true nor false (without specifying who Tom is) - it is an expression which could be true or false, depending on who we mean when we say "Tom".

On the other hand, the statement 1+1=0 is a false. Analogously, the statement "If Tom has a cat then Tom has no cats" is false, no matter who we mean when we say "Tom".

In this case, 1+1=1 is the true statement. Its dual is 0.0=0, which is also a true statement.

$\endgroup$
4
  • $\begingroup$ Thanks for your explanation. I am no mathematician, so I need to have things explained in layman terms. So from your answer I understood that 1+1 is not a statement, so it makes no sense to compare its value to 0.0 and to talk about duality. On the other hand 1+1=1 and 0.0=0 are two true statements, so the duality principle is valid. Is this the correct understanding ? $\endgroup$
    – jcxz
    Jun 1, 2013 at 12:33
  • $\begingroup$ 1 + 1 is indeed true, as a boolean statement, as the truth values are given [as true = 1], and + is defined as the boolean "or", so "true or true" has a truth value, and that happens to be true, just as 0.0 reads "false and false" which is false, so "false and false = false" is, as a whole, a true boolean statement. $\endgroup$
    – amWhy
    Jun 1, 2013 at 12:40
  • 1
    $\begingroup$ An expression has a value, 1 or 0, but it is not a statement. As your question demonstrates, the duality principle only applies to statements, that is, things which contain an = sign. Perhaps it helps to think about 2+2 using normal addition. This clearly evaluates to 4, but we have no notion of whether it's true or not. However, 2+2=4 is clearly true. $\endgroup$ Jun 1, 2013 at 15:36
  • $\begingroup$ Yes, indeed. I think I was just misunderstanding what you wrote: and the case here is a bit more complicated than addiction of integers, because here $1$ is identically true, and $0$ is identically false. $\endgroup$
    – amWhy
    Jun 1, 2013 at 17:54
1
$\begingroup$

The statement is not 1+1 but rather 1+1=1.

What the duality principle says is that "if you exchange every symbol by its dual in a formula you get the dual result".

$\endgroup$
1
$\begingroup$

The difference between an expression and a statement is that statement is like an equation and expression is like variable.

"Duality of a statement is true" - What this means is that when you equate the dual of both the expressions in LHS and RHS of a statement(equation), the statement still makes sense.

Observe that in your example both 1+1 = 1 & 0.0 = 0 ,which are dual to each other, are true statements.

$\endgroup$
0
$\begingroup$

The dual of $1.0$ which is $0+1$, will be obtained by interchanging $AND$ to $OR$ and $1$'s to $0$'s. The result of those function needn't be same.The results will be opposite to each other $1.0=0;0+1=1$, but they are the dual form of the single function $1.0$.
For example, the dual function of $XNOR$ is $XOR$.

$\endgroup$
-1
$\begingroup$

In my humble opinion, it is more helpful to describe the duality relationship in terms of variables. Example: Expression: x + x = x Dual of expression: x * x = x x could be 1 or 0. In all iterations, the expressions are correct. Good Luck!

$\endgroup$
1
  • $\begingroup$ Shouldn't the dual of $ x + x = x$ be $x' \cdot x'=x'$? $\endgroup$
    – Ardent
    Sep 20, 2019 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.