Question

Prove that \[\left( {100 - 100} \right)\]is divided by \[\left( {100 - 100} \right)\]is equal to 2.

Answer 1

Hint: Normally all will tries to proof it by solving the dividends like \[\dfrac{{\left( {100 - 100} \right)}}{{\left( {100 - 100} \right)}} = \dfrac{0}{0}\]; the result is undefined. Division by zero is undefined. Since any number multiplied by zero is zero, the expression \[\dfrac{0}{0}\] has no defined value , when it is in the form of limit , it is an indeterminate form. Therefore, if we think in a normal sense the answer comes non-defined instead of 2. In this type of sum, we should think in different ways. We must use a formula of algebra to prove this sum. We know \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].This formula is very important to solve this sum. Then we need to take common denominators. After doing some more steps we will get the desired result.

Complete step by step answer:
To prove :- \[\dfrac{{\left( {100 - 100} \right)}}{{\left( {100 - 100} \right)}}\]\[ = 2\]
Solving L.H.S
\[\dfrac{{\left( {100 - 100} \right)}}{{\left( {100 - 100} \right)}}\] can be written as \[\dfrac{{\left( {{{10}^2} - {{10}^2}} \right)}}{{\left( {100 - 100} \right)}}\]
Since we know \[{10^2} = 100\] i.e \[10 \times 10 = 100\].
Further \[\dfrac{{\left( {{{10}^2} - {{10}^2}} \right)}}{{\left( {100 - 100} \right)}}\] can be written as \[\dfrac{{\left( {{{10}^2} - {{10}^2}} \right)}}{{\left( {10 \times 10 - 10 \times 10} \right)}}\].
\[ \Rightarrow \dfrac{{\left( {10 + 10} \right)\left( {10 - 10} \right)}}{{\left( {10 \times 10 - 10 \times 10} \right)}}\]
Using the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] in the numerator .

Taking \[10\] common from the denominator ; we get
\[ \Rightarrow \dfrac{{\left( {10 + 10} \right)\left( {10 - 10} \right)}}{{10\left( {10 - 10} \right)}}\]
We know when we have same factor in numerator as well as in denominator we can cancel it out ; after cancelling \[\left( {10 - 10} \right)\]both from numerator and denominator , we get,
\[ \Rightarrow \dfrac{{\left( {10 + 10} \right)}}{{10}}\]
As \[10 + 10 = 20\]; Putting the value , we get
\[ \Rightarrow \dfrac{{20}}{{10}}\], which can be written as,
\[ \Rightarrow \dfrac{{2 \times 10}}{{10}}\]
since \[2 \times 10 = 20\]
Now again cancelling out the same factor i.e \[10\] which is present in both denominator and numerator , we get
\[ \Rightarrow \dfrac{2}{1}\]= \[2\];
Therefore L.H.S = \[2\]
As R.H.S = \[2\];
Hence L.H.S =R.H. S

Therefore \[\dfrac{{\left( {100 - 100} \right)}}{{\left( {100 - 100} \right)}}\]\[ = 2\] ( Proved ).

Note: For this kind of unique type of sum, we need to have a smart approach towards it and must think out of box to get the solution. So, first find the square root, and do not cancel any common factor in between the solution as it leads you to the wrong answer.Always cancel the common factors at the end of the solution when there is no chance of further calculation. Also remember most of the times to solve such type of problems, the main formula used is \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] only as it involves the difference of two square terms.