Question

Is the following true for any value of a and b?
\[\dfrac{1}{ab}=\dfrac{1}{a}-\dfrac{1}{b}\]

Answer 1

Hint: To solve the given question, we will take the LCM of the numbers on the right-hand side and simplify them. Then we will multiply both the sides of the equation by ab. From here, we will get a relation in a and b. Now, we will put different values of a and b and then we will check whether the equation which we have obtained satisfies all the pairs of (a, b) or not. On this basis, we will answer the question.

Complete step-by-step answer:
To start with, we are given two fractions on the left-hand side of the equation. So, the first thing which we are going to do is to take the LCM of the terms on the left-hand side. Thus, we have the following equation.
\[\dfrac{1}{ab}=\dfrac{1}{a}-\dfrac{1}{b}\]
\[\Rightarrow \dfrac{1}{ab}=\dfrac{1\left( b \right)-1\left( a \right)}{ab}\]
On simplifying this, we will get the following equation
\[\Rightarrow \dfrac{1}{ab}=\dfrac{b-a}{ab}\]
Now, we will multiply both sides of the equation by ab. Thus, we will get the following equation.
\[\Rightarrow ab\times \dfrac{1}{ab}=ab\times \dfrac{b-a}{ab}\]
On canceling the terms, we will get the following equation.
\[\Rightarrow 1=b-a\]
\[\Rightarrow b-a=1\]
Now, we will put random values of a and b and then check whether they are equal. Let us assume that a = 4 and b = 5. Thus, we have,
\[\Rightarrow 5-4=1\]
Here, LHS = RHS. So, for these values of a and b, the equation is satisfied. Now, let us assume that b = 3 and a = 5. Thus, we will get,
\[\Rightarrow 3-5=1\]
\[\Rightarrow -2=1\]
In this case, \[LHS\ne RHS.\]
So, we can say that the relation \[\dfrac{1}{a}-\dfrac{1}{b}\] does not satisfy all values of a and b.

Note: We can directly come to the conclusion without even simplifying the given equation because we know that when we will put a = 0 and b = 0 in the above equation, the resultant equation will become
\[\dfrac{1}{0\times 0}=\dfrac{1}{0}-\dfrac{1}{0}\]
\[\Rightarrow \infty =\infty -\infty \]
This is a wrong equation so we can say that a and b cannot have any random values.