Is ZERO even??

[NP-complete]

Wrong. We had all that. Some invented earlier by Greeks and such and some invented by us, Indians.

Well not exactly. Back then people accepted the concept of zero. They accepted the concept of negative integers, which comes from the idea of zero (integer and its corresponding negative integer add to give zero, kind of like matter/anti-matter). They were familiar with the idea of rational numbers and had used simple fractions. They even accepted the concept of irrational numbers. But for all practical purposes they did all their maths with positive integers only. Others were incorporated into mathematical calculations by arabs later in the middle ages.

Challenge here is generating series of integral solutions, and coming up with generalized formula. This is the same person (Brahmagupta) who invented zero as we know till now.

I consider solving these quadratic diophantine equations advanced number theory.

 

[Levina]

Please define characteristics of both zero....is it zero of

Indian or Pakistani
Aam aadmi of politician
Public sector or private sector
Socialist or capitalist
insurgent or Army
Republican or Democrat

Zero mane annda nothing more nothing less..

 

[liall]

English thi!!

Now prove 1=0..I am curious.


I use French when I have to use some blessed words.

Btw I got a puzzle for you. You can use 0 a total of 6 times. Use any mathematical operators and give me 22. I was asked this in an interview lol

 

[Levina]

Btw I got a puzzle for you. You can use 0 a total of 6 times. Use any mathematical operators and give me 22. I was asked this in an interview lol

That's interesting!!
Go on...

 

[liall]

That's interesting!!
Go on...

I meant for you to solve it. That was the whole puzzle right there

 

[Biplab Bijay]

x = x
=> x2 = x2
=> x2 -x2 = x2 - x2
=> x (x - x) = (x + x) ( x - x)
=> x = 2x
=> 1 = 2
=> 0 = 1


@levina

English thi!!

Now prove 1=0..I am curious.


I use French when I have to use some blessed words.

x = x
=> x2 = x2
=> x2 -x2 = x2 - x2
=> x (x - x) = (x + x) ( x - x)
=> x = 2x
=> 1 = 2
=> 0 = 1

 

[Jf Thunder]

Maths is something that you carry out on daily basis...from the number of almonds you eat in the morning to the addition/substraction you do while bargaining at the market.
Numbers are fun!!!

ok, i hate you now

 

[Levina]

x = x
=> x2 = x2
=> x2 -x2 = x2 - x2
=> x (x - x) = (x + x) ( x - x)
=> x = 2x
=> 1 = 2
=> 0 = 1


@levina




x = x
=> x2 = x2
=> x2 -x2 = x2 - x2
=> x (x - x) = (x + x) ( x - x)
=> x = 2x
=> 1 = 2
=> 0 = 1

Smart!
But that proves 1=2 not 0=1.
And in the Third step both the sides are not equal. What do we do about it??

I meant for you to solve it. That was the whole puzzle right there

Lol
I'm not so smart...I give up.

ok, i hate you now

 

[Jf Thunder]

hain ? ha in hain hain ?

hain?

Smart!
But that proves 1=2 not 0=1.
And in the Third step both the sides are not equal. What do we do about it??

Lol
I'm not so smart...I give up.

awkward

 

[Biplab Bijay]

Smart!
But that proves 1=2 not 0=1.
And in the Third step both the sides are not equal. What do we do about it??

Lol
I'm not so smart...I give up.

why third steps are not equal.

 

[NP-complete]

x = x
=> x2 = x2
=> x2 -x2 = x2 - x2
=> x (x - x) = (x + x) ( x - x)
=> x = 2x
=> 1 = 2
=> 0 = 1

O bhai, here you divided the equation by zero (x-x). Division by zero is not allowed.

 

[Biplab Bijay]

O bhai, here you divided the equation by zero (x-x). Division by zero is not allowed.

X is unknown. In LHS i have taken x as common. In RHS I use the formula a2-b2

 

[Levina]

why third steps are not equal.

the third step is this
x (x - x) = (x + x) ( x - x)

there's nothing wrong with how you've proceeded but if at this step you remove (x-x) from both the sides then we are left wit an excess X on rhs.
I mean how is this possible logically??

 

[NP-complete]

X is unknown. In LHS i have taken x as common. In RHS I use the formula a2-b2

x is unknown but x-x is known. It is zero.
How did, after x(x-x) = (x+x)(x-x), you get x = 2x ?

 

[Biplab Bijay]

the third step is this
x (x - x) = (x + x) ( x - x)

there's nothing wrong with how you've proceeded but if at this step you remove (x-x) from both the sides then we are left wit an excess X on rhs.
I mean how is this possible logically??

I used cancellation law

if a*x = b*x then a = b for all x belongs to R.

x is unknown but x-x is known. It is zero.
How did, after x(x-x) = (x+x)(x-x), you get x = 2x ?

Cancellation law.

I used cancellation law

if a*x = b*x then a = b for all x belongs to R.