Question

Prove the following mathematically :-

$\bigstar \large \boxed{ \bf \frac{0}{0} = 2} \bigstar$

Do it again ._.

Have a great answering time xD ✌​

Answer 1

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Here, we'll have to prove that $\dfrac{0}{0}=2.$

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For this, we can say that :

$\qquad \dashrightarrow \: \dfrac{100 - 100}{100 - 100} = 0$

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The number $100$ can be written as $10\times10$ :

$\qquad \dashrightarrow \: \dfrac{(10 \times 10) - (10 \times 10)}{(10 \times 10) - (10 \times 10)}$

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Also, $(10\times10)$ can be written as $10^2$ in the numerator :

$\qquad \dashrightarrow \: \dfrac{ ( {10)}^{2} - ( {10)}^{2} }{(10 \times 10) - (10 \times 10)}$

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In the denominator, the number can be written as $10(10-10)$ by removing out the common term :

$\qquad \dashrightarrow \: \dfrac{( {10)}^{2} - ( {10)}^{2} }{10(10 - 10)}$

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Using $(a-b)(a+b)=a^2-b^2$ in the numerator :

$\qquad \dashrightarrow \: \dfrac{(10 + 10)(10 - 10)}{10(10 - 10)}$

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The term $(10-10)$ gets cancelled our with the term $(10-10)$ in the denominator :

$\qquad \dashrightarrow \: \dfrac{(10 + 10) \: \cancel{(10 - 10)}}{10 \: \cancel{(10 - 10)}}$

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The fraction now looks like this :

$\qquad \dashrightarrow \: \dfrac{10 + 10}{10}$

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Evaluating the fraction further :

$\qquad \dashrightarrow \: \cancel{\dfrac{20}{10} }$

$\qquad \dashrightarrow \: \dfrac{0}{0} = 2$

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Henceforth, proved mathematically!

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Answer 2

Given :-

0/0 = 2

To Prove :-

The Given

Solution :-

It is only a trick Surely some others will accept this But I can never accept this . We can't accept this mathematically because we knows that 0/0 is indeterminate. I will also explain why it is not accepted mathematically.

=> To prove : 0/0 = 2

Let , LHS i.e 0/0

=> 0/0

=> 4 - 4/4 - 4

=> ( 2 )² - ( 2 )²/ 2 ( 2 - 2 )

=> ( 2 + 2 ) ( 2 - 2 )/2 ( 2 - 2 )

{ Because a² - b² = ( a + b ) ( a - b ) }

=> 4/2

=> 2

Henceforth , Done :)

Now , you can see that here we used a² - b² formula to which we are aware from very starting . But this is wrong , because a² - b² is only applicable if and only a is not equal to b .

And why this condition is essential let us understand this also i.e

If a = b , Then ,

a² - b²

=> ( a + b ) ( a - b )

=> Put a = b ,

=> ( b + b ) ( b - b )

=> 2b × 0

=> 0

But this is not possible because. , ( 7 )² - ( 6 )² is not equal to 0 and if this becomes 0 then this formula has no mean .But this formula is the base of many theorems and formulae . Thus , a can't equal to b.

You should note that we can also prove 0/0 to 1 But this is also not supported mathematically . Let us see how ,

=> To Prove : 0/0 = 1

=> Let LHS i.e 0/0

=> 4 - 4/4 - 4

=> On Rationalising the denominator we get ,

$=> \: \frac{4 - 4}{4 - 4} \times \frac{4 + 4}{4 + 4}$

$=> \: \frac{ {(4)}^{2} - {(4)}^{2} }{(4 + 4)(4 - 4)}$

Applying a² - b² = ( a + b ) ( a - b ) we get ,

$=>\frac{(4 + 4)(4 - 4)}{(4 + 4)(4 - 4)}$

=> 1/1 = 1

Henceforth , Done :)