math guru over to you.

Assuming A to be 2 and be to be 1

2-2 = 1-1 = 1=0

So A is not equal to B

A=A and A does not equal to B

Bigredmachine:
No A is not B...

elaborate please with mathematical prove.

Iyaebe:
A=A and A does not equal to B

elaborate please with mathematical prove

silibaba:
elaborate please with mathematical prove

A-A=B-B A=B-B+A so the - and + B cancel themselves out and you'll be left with A=A

A-A= B-B
A-A-B+B =0
(A-A)-(B-B) =0
A(1-1)-B(1-1) =0
(A-B)(1-1) = 0
So, (A-B) =0 and (1-1)= 0
A-B = 0 and 1-1 = 0
So, A =B

A-A = B-B
Simultaneous equation says solve A-A first.

(A-A) = 0

(B-B) = 0

Therefore A=B is 0

ChybuzzDD:
A-A= B-B
A-A-B+B =0
(A-A)-(B-B) =0
A(1-1)-B(1-1) =0
(A-B)(1-1) = 0
So, (A-B) =0 and (1-1)= 0
A-B = 0 and 1-1 = 0
So, A =B

A=B? I don't think so.

You can't conclude that A = B except further conditions are stated.

It can hold true when further conditions are stated. Therefore it is mathematically wrong to conclude that A = B from just one condition.

You will need an extra condition to make the equation mathematically feasible.

You have one equation and two unknowns. It can't work. We need one more equation.

I need to know this Maths like u Gurus up there.

.

No.

3 – 3 = 7 – 7 & 2 – 2 = 0 – 0.

This doesn't mean 3 = 7 and 2 = 0.

Obviously, A=A and B=B (assumption).

Hence, since A-A=0, and B-B=0, you can clearly see why A-A=B-B.

However, let A=7 and B=5.

Then, A-A=7-7=0.

Also, B-B=5-5=0.

Interpretation: if you subtract anything from itself, you will ALWAYS get 0, irrespective of the values of that thing.

Therefore, from the above deductions, we can objectively conclude that A-A=B-B does not necessarily mean that A=B

let me try small

If a - a = b - b then does a = b?
No.

This confusion arises because of missing the fact that while applying functions on the two sides of an equation, we are only allowed to move “forwards” and not “backwards”.

Let us say we have an equation that looks like

LHS=RHS

If we have any function f that can be applied on the domain, we can now say

f(LHS)=f(RHS)

But moving in the reverse direction is only allowed if f is a one-one function.[1]

As an example, take f(x)=x2 . This is not a one-one function because (−x)2=x2 . Hence

x=y⟹x2=y2

But

x2=y2⇏x=y

Because it might be that x=−y .

In this question, we just have the function f(x)=0 . Since it is clearly not one-one, we cannot reverse the step.

just hope am right