Question

Mathematically prove followings:-

1. $\bold\green{\dfrac{0}{0} }$ = $\huge{\boxed{\bold{2}}}$

2. $\bold\green{\sqrt{2} \: is\:irrational}$

Quality Answer needed!!!​

Answer 1

Question (1) :--

Prove that 0 divided by 0 is Equal to 2.

First of all, if we divide anything with Zero, we Always gets a undefined value .

So, Mathematically its wrong .

But , To play with maths (Just for Fun, Lets Do it now ) .

Solution :--

To Prove :- 0/0 = 2 .

Taking LHS ,

→ 0/0

we can write it as

→ (100 - 100) / (100 - 100) ( As value of both are 0) .

Now, we have 10² = 100 , So, we can write ,

→ [ (10)² - (10)² ] / [ 100 - 100 ]

Taking 10 common From denominator part and Using a²-b² = (a+b)(a-b) in Numerator , we get,

→ [(10+10)(10-10)] / 10(10-10)

Now, (10-10) will be cancel From both sides .,

So, we have ,

→ (10+10)/10

→ 20/10

→ 2 = RHS .

✪✪ Hence Proved ✪✪

[ Note :- Only For Fun part ]

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Question (2) :-- Prove that √2 is a irrational Number . ?

Irrational Number :- A number which cant be written in the form of p/q, (where q ≠ 0 and p and q are co-prime numbers ) .

Solution :--

First of all lets assume that √2 is a rational Number .

So,

→ √2 = p/q . (where q ≠ 0 and p and q are co-prime numbers )

Now, Squaring both sides we get,

→ 2 = (p²/q²)

→ 2q² = p² --------------------- Equation (1)

With this we can say that, 2 divides p or factor of p (as p is a prime number ) .

Now, Lets assume p = 2c. ( is p is a factor of 2) .

putting this value in Equation (1) now, we get,

→ 2q² = (2c)²

→ 2q² = 4c²

→ q² = 2c²

Here , now can also see that, 2 divides q also . ( or Factor of q also).

Or, we can now say that, both p and q have a common Factor other than 1 now .....

This conclude that, our Assumption of √2 is a rational Number is wrong ..

So, we can say that √2 is an irrational Number.

Answer 2

Solution 1).

To prove 0/0 = 2

=> (100-100)/(100-100)

=> (10²-10²)/10(10-10)

using a² - b² = (a+b)(a-b)

=> [(10+10)(10-10)]/10(10-10)

=> 20/10

=> 2

•°• 0/0 = 2

Hence proved!

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Solution 2).

Let √2 is a rational number.

√2 = p/q ( where p and q are co prime numbers )

Squaring both sides,

=> 2 = p²/q²

=> 2q² = p² _______________(1)

From (1) is it clear that, p² is divisible by 2 and hence p is a factor of 2.

Let p = 2m [ where m is any whole number ]

squaring both sides,

=> p² = (2m)²

=> 2q² = 4m² [ From (1) ]

=> q² = 2m²

•°• q² is divisible by 2 and hence q is factor of 2.

But it isn't possible. Therefore our supposition is wrong.

•°• √2 is irrational number.

Hence proved!