Is ZERO even??

[Levina]

I used cancellation law

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


Cancellation law.

I used cancellation law

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

naa
I understood the step...but i'm wondering how can both the sides be equal at that stage??

 

[Biplab Bijay]

naa
I understood the step...but i'm wondering how can both the sides be equal at that stage??

There in LHS I have taken x as common and RHS I have used a2 - b2 = ( a + b) ( a - b). I did not violate any of the mathematical theorems . he he he

 

[Levina]

There in LHS I have taken x as common and RHS I have used a2 - b2 = ( a + b) ( a - b). I did not violate any of the mathematical theorems . he he he

I know!!

your steps were all fine.
but then how is it logically possible?

 

[NP-complete]

There is no such thing as cancellation law. We cannot cancel anything from both sides of the equation. If we want to remove a factor that is common on both sides, we can divide the equation by that factor and it will get divided on both sides. One will be left. Cancellation terminology is a shortcut. These are the real steps,

ab = ac
(ab)/a = (ac)/a
(1)(b) = (1)(c)
b = c

You divided the equation from x-x, which is zero. Division by zero is not allowed.

 

[Biplab Bijay]

I know!!

your steps were all fine.
but then how is it logically possible?

You tell me. if you find anything wrong then prove it. he he he he he he he.

 

[Levina]

You tell me. if you find anything wrong then prove it. he he he he he he he.

let me watch @NP-complete ...
he's right..you divided it by x-x which is zero.hehe

 

[Biplab Bijay]

There is no such thing as cancellation law. We cannot cancel anything from both sides of the equation. If we want to remove a factor that is common on both sides, we can divide the equation by that factor and it will get divided on both sides. One will be left. Cancellation terminology is a shortcut. These are the real steps,

ab = ac
(ab)/a = (ac)/a
(1)(b) = (1)(c)
b = c

You divided the equation from x-x, which is zero. Division by zero is not allowed.

There is no such thing as cancellation law ? Are you sure ?

You must read group theory. Ok ? may be Algebra by I. N. Herstein or A descent group theory book. Its a group property.

let me watch @NP-complete ...
he's right..you divided it by x-x which is zero.hehe

Where , when ?

 

[Levina]

There is no such thing as cancellation law ? Are you sure ?

You must read group theory. Ok ? may be Algebra by I. N. Herstein or A descent group theory book. Its a group property.


Where , when ?

hmmmm

see at this step
x (x - x) = (x + x) ( x - x)
both sides were expanded and (x-x) was common on both sides.
So the logical next step is
x (0) = (x+x) (0)
which would mean zero in the end.
Instead you cancelled out (x-x) on both the sides.

 

[Biplab Bijay]

let me watch @NP-complete ...
he's right..you divided it by x-x which is zero.hehe

By the way I know where the problem is. if any one can point out, then a chocolate for him/her. I have already given a hint.

hmmmm

see at this step
x (x - x) = (x + x) ( x - x)
both sides were expanded and (x-x) was common on both sides.
So the logical next step is
x (0) = (x+x) (0)
which would mean zero in the end.
Instead you cancelled out (x-x) on both the sides.

Why should I go x-x = 0, I will just cancel them before.

 

[NP-complete]

There is no such thing as cancellation law ? Are you sure ?

You must read group theory. Ok ? may be Algebra by I. N. Herstein or A descent group theory book. Its a group property.

If you want to use group theory terminology then you should know that zero doesnt have either left cancellation property or right cancellation property.

 

[Levina]

By the way I know where the problem is. if any one can point out, then a chocolate for him/her. I have already given a hint.


Why should I go x-x = 0, I will just cancel them before.

my mind is saturated now!!
ab bataoo

 

[Biplab Bijay]

my mind is saturated now!!
ab bataoo

ha ha ha. This is from group theory.
suppose a, b, c belong to a group G under operation *( here * is multiplication)

Then a, b, c, all of them must have inverses in that group.

now a*c = b * c => a = b for all c belongs to G and there must be an inverse of C

Now suppose x is an element of G under * ( here * is multiplication)
so the inverse is 1/x
x2 is an element of G.

But again if a is an element of G under multiplication then a - a is not an element of G under multiplication.
so x2 - x2 is not an element of G under multiplication.
so we can not do x2 - x2 = x2 - x2 under G as this is not an element of G.

he he he he.

 

[Spade]

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

Doesn't matter if X is known or unknown. What matters is whether it is same or not (whether X=X or not). If it is same, you are dividing by 0 (as already pointed out, X-X=0) which is not allowed. If it is not same(X=/=X), you can't use X for both, you need to use something else to denote 2 different unknowns and X+X=2X also becomes invalid.

 

[Biplab Bijay]

Doesn't matter if X is known or unknown. What matters is whether it is same or not (whether X=X or not). If it is same, you are dividing by 0 (as already pointed out, X-X=0) which is not allowed. If it is not same(X=/=X), you can't use X for both, you need to use something else to denote 2 different unknowns and X+X=2X also becomes invalid.

You can not say it simply. You have to use group properties. he he he.

 

[Spade]

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.

I consider solving these quadratic diophantine equations advanced number theory.

Wrong again. It was Brahmagupta who gave practical operations with negative numbers(- X - = +, + X - = -). He said root of any number had 2 answers, positive and negative. It was Arabs and Europeans who thought this was absurd, until 1700s!! Alberuni accepted that he go the idea of -ve numbers from Brahmagupta but said it was meaningless. Regarding diophantine equations is a 20th century invention. A form of it was invented solved by Bhaskara 2 . around 1200 CE (Chakravala method). The equation 61x^2+1=y^2 was solved by europeans only in 1800...

You can not say it simply. You have to use group properties. he he he.

why?