Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

With additive inverse operations verify that: -(-x) = x for –
(i) x = $\dfrac{11}{15}$
(ii) x = $-\dfrac{13}{17}$

seo-qna
Last updated date: 16th Apr 2024
Total views: 420.3k
Views today: 6.20k
MVSAT 2024
Answer
VerifiedVerified
420.3k+ views
Hint: To solve this problem, we must be aware about the basic arithmetic (in this case additive inverse) operations. Additive inverse of the number is basically negative. We use this property to verify this problem.
Before solving the problem, we will briefly try to understand the basics of additive inverse. Basically, the additive inverse of a number a is the number that, when added to a, yields zero. We understand this through an example. Let’s take a rational number a = 4. Now, to find the additive inverse of a, we have,
a+x=0
4+x=0
x = -4

Complete step-by-step answer:
Thus, the additive inverse of a=4 is -4. Now, we begin to solve the problem in hand. We have,
(i)$\dfrac{11}{15}$
We follow the above given procedure to find the additive inverse for part (i). Let x=$\dfrac{11}{15}$ and a be the additive inverse of x. Thus, we have,
a+x=0
$\dfrac{11}{15}$+a=0
a= -$\dfrac{11}{15}$
Thus, a = -x = -$\dfrac{11}{15}$.
Now, to find -(-x), we now find the additive inverse of –x (that is -$\dfrac{11}{15}$), we let the additive inverse be b. Thus,
b+ (-x) =0
b+$\left( -\dfrac{11}{15} \right)$=0
b = $\dfrac{11}{15}$= - (-x)
Which is the same as x, thus, x = - (-x).

 (ii) $-\dfrac{13}{17}$
We again follow the above given procedure to find the additive inverse for part (ii). Let x=$-\dfrac{13}{17}$ and a be the additive inverse of x. Thus, we have,
a+x=0
$-\dfrac{13}{17}$+a=0
a= $\dfrac{13}{17}$
Thus, a = -x = $\dfrac{13}{17}$.
Now, to find -(-x), we now find the additive inverse of –x (that is $\dfrac{13}{17}$), we let the additive inverse be b. Thus,
b+ (-x) =0
b+$\left( \dfrac{13}{17} \right)$=0
b = $-\dfrac{13}{17}$= - (-x)
Which is the same as x, thus, x = - (-x).

Note: The use of additive inverse is useful in solving algebraic terms, where we have to cancel the terms by making use of additive inverse property. In short, we should remember that the additive inverse of a number is simply that number multiplied by (-1).