Question

Show that the sum of an integer and its additive inverse is always zero.

Answer 1

Hint: Additive inverse of a number $a$, refers to the number which when added to the number $a$ gives zero. The additive inverse of $a$ is represented by $ - a$.
If any statement is shown valid for any arbitrary point of the set of integers, this means that the statement is true for every integer.

Complete step-by-step solution:
We are required to show that the sum of an integer and its additive inverse is always zero.
Let us assume any arbitrary integer, say $a$.
Now, we need to determine the additive inverse of this number $a$. Since the additive inverse of any number $a$ is represented by $ - a$.
So we need to show that the addition of this arbitrary integer $a$ to its additive inverse always gives zero as its solution.
We have,
$a + ( - a)$
We are required to solve the above terms and show that the result is zero.
Let us assume that the result is not zero and some integer, say $b \ne 0$.
Then, $a + ( - a) = b$
Now, add $a$ on both sides of the above equation, we get
$a + ( - a) + a = b + a$
Using associative property
$a + ( - a + a) = b + a$
Since, $ - a + a = 0$
$a + 0 = b + a$
$a = b + a$
As, we have taken $b$ to be a non-negative integer, now if we add any non-negative integer to $a$, so that the result is also the same, that is $a$. This cannot be possible unless $b$ is zero.
Hence, we get that
$a + ( - a) = 0$
Since,$a$ is an arbitrary integer, so it is true for all the integers.
Thus, the sum of an integer and its additive inverse is always zero.

Note: To show any statement to be true, an indirect method can be used. This approach works when we assume that the given statement is not true and proceed with the usual operations to achieve a statement which is absurd and is a point of contradiction.