Non-Zero and Zero Products
Non-Zero and Zero Products
Decimal methods for multiplication
implies the product of two whole number 123 and 583 is nonzero.
The same or similar decimal methods imply the product of nonzero
decimals say 0.0123 = 123 &\divide; 100 and 5.83 = 10 ÷ 10 is given by shifting of the
decimal point in the non-zero product of 123 and 583. Prefixing the
factors with signs will give a signed nonzero number for the product.
Non-Zero Product Law For Multiplication - Direct Form
Axiom: If a and b are nonzero real numbers then
\[ a \times b \ne 0 \]
This law is logically equivalent to saying that
computation rule \[f(a,b) = a \times b \]
will give nonzero results when a and b are nonzero.
What happens when products are zero?
Multiplication by zero or taking zero times another number gives zero. Moreover,
zero times zero is zero. That follows from primary school experience in learning to count.
Now the implication rule in the axiom
If a and b are nonzero real numbers then
$a \times b \ne 0 $
would be disobeyed if we had numbers or number-valued expressions a and b for which
$ab = 0$ and both a and b are nonzero. So for the axiom not to be disobeyed, we must
have the following occur.
If a and b are real numbers with their product
$a \times b = 0 $ then at least one of the factors a and b is zero.
This implication rule is not used directly in computation, but it is used
in finding solutions to equations. Examples follow. Assume the letters in them
stand for real numbers.
If $ 5x=0 $ then x = 0 since $5\ne 0$
If $ (x-4)3 = 0 $ then x - 3 = 0 since $4 \ne 0.$ Therefore
(x-4)3 = 0 has only one solution x = 3.
If $ 0 = x(x-5) $ then at least one of the factors x and x-5 must be
zero. By inspection x = 0 and x = 5 from the case where x - 5 = 0 both give
solutions of the equation.
If $(x-3)(x-1)= 0 $ then instead of saying at least one of the factors
x-3 and x-1 is zero, we write Therefore (i) x- 3 = 0 or (ii)
x -1 =0. Here the condition x - 3 = 0 is equivalent to x = 3 while the condition
x - 1 = 0 is equivalent to x = 1. Whence we conclude the equation $(x-3)(x-1)= 0 $
requires x = 3 or x = 1. To check, we observe product is zero when x = 3 and when x =1.
Note saying x = 3 or x = 1 does not meand x has
the both values at the same moment. That is like saying when
a coin lands, it the top size will be a head or a tails.
If $y(y-10) = 0 $ and there is a further requirment that y be less than
10 then the equality $y(y-10) = 0 $ still implies y = 0 or y = 10, but the solution
is not y = 10 because of the extra condition.
Extended Zero Product Law
If a product of real numbers is zero then a
then at least one of the factors in the product is zero.
In equation solving situation, this extended product law implies
many possibilities. However, additional information may be present
to eliminate some of the possibilities or all. Watch for that.
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