Question

Multiplication of a negative number for even time gives a _______ result
 $
  (a)\,\,negative \\
  (b)\,\,0 \\
  (c)\,\,positive \\
  (d)\,\,none \\
  $

Answer 1

Hint: We know that under multiplication of two negative terms their minus signs get cancelled and hence proceed in the same manner if there are even negative terms then writing them in a pair results cancelation of negative signs and therefore the result will be a positive number.

Complete step-by-step answer:
To discuss the required solution of the given problem. We take different examples to find the correct answer. Which are as follows:
In the first case we consider a product of two negative numbers say (-A) and (-B).
We write product as $ \left( { - A} \right) \times \left( { - B} \right) $
\[ \Rightarrow \left\{ {\not - A \times \not - B} \right\},\,\,\,cancellation\,\,of\,\,\,negative\,\,sign\]. We have
 $ A \times B $ Which will definitely be a positive number.
In the second case we consider a set of four numbers (can’t take 3, as asked to take product of even) say (-A), (-B),(C) and (D).
Writing in product form. We have,
\[\left\{ {\left( { - A} \right) \times \left( { - B} \right) \times \left( { - C} \right) \times \left( { - D} \right)} \right\}\]
Writing in pair form we have,
 $
  \left\{ {\left( { - A \times - B} \right) \times \left( { - C \times - D} \right)} \right\} \\
  \left\{ {\left( {\not - A \times \not - B} \right) \times \left( {\not - C \times \not - D} \right)} \right\} \\
   \Rightarrow \left\{ {\left( {A \times B} \right) \times \left( {C \times D} \right)} \right\} \\
   \Rightarrow \left( {A \times B \times C \times D} \right) \;
  $
Clearly products of four even negative numbers end with a positive number.
Now, if we consider products of n even negative terms. For this let n negative even numbers are:
 $ - {x_1}, - {x_2}, - {x_3}, - {x_4},.............. - {x_{2n}} $
Writing in product form. We have
 $ - {x_1} \times - {x_2} \times - {x_3} \times - {x_4}.............. \times - {x_{2n}} $ $ $
Writing them in pair form. There will be a complete pair formed with no single term left out as the number of terms are even. We have
 $ \left\{ {\left( { - {x_1} \times - {x_2}} \right) \times \left( { - {x_3} \times - {x_4}} \right) \times .............\left( { - {x_{2n - 1}} \times - {x_{2n}}} \right)} \right\} $
Cancelling minus signs in pairs. We have,
 $ \left\{ {\left( {\not - {x_1} \times \not - {x_2}} \right) \times \left( {\not - {x_3} \times \not - {x_4}} \right) \times .............\left( {\not - {x_{2n - 1}} \times \not - {x_{2n}}} \right)} \right\} $
 $ \Rightarrow \left\{ {\left( {{x_1} \times {x_2}} \right) \times \left( {{x_3} \times {x_4}} \right)......... \times \left( {{x_{2n - 1}} \times {x_{2n}}} \right)} \right\} $
Clearly from above we see that products of n even negative terms are also positive.
Hence, from above we say that a product of even numbers of negative terms always ends positive irrespective of how many even numbers there are in the product.
So, the correct answer is “Option C”.

Note: Solution of the given problem can also be discussed in other ways. In this way we first separate all negative signs from given numbers writing as a product of $ - 1 \times - 1 \times - 1 \times - 1....... $ Now, by exponent law we can write $ - 1 \times - 1 \times - 1 \times - 1....... = \,\,{( - 1)^{2n}} $ . Here,$2^n$ is taken as a number of terms in a product are always even. And we know that an exponent negative base with even power results positively. Therefore, the result of even negative terms always ends in a positive number.