You were correct till (a-b)(a+b) =b(a-b), keeping in mind that a=b=1.
But when you solve the equations, you always place a condition, and in this case, for you to transform (a+b) (a-b) =b(a-b) to (a+b)(a-b)/(a-b) = b, the condition is that a-b= not 0, so a= not b.
But you have a=b=1(or even any other number), so you cannot do that transformation.
Replying to:ArtelionThis is why you can't do x÷0 where x=lR.
You were correct
But yeah, i understand what you mean. When you are coding and you get an error, sometimes it can be really hard to find the error, specially when it is a long code.
I always take a break when i am coding long files to check if there are any errors, and if there are none, i save them as a separate backup file, like that it will be easier to solve any mistake you encounter.
You never had a=1 in the equation. So it doesn’t even matter.
This is why you can’t do x÷0 where x=lR.
You were correct till (a-b)(a+b) =b(a-b), keeping in mind that a=b=1.
But when you solve the equations, you always place a condition, and in this case, for you to transform (a+b) (a-b) =b(a-b) to (a+b)(a-b)/(a-b) = b, the condition is that a-b= not 0, so a= not b.
But you have a=b=1(or even any other number), so you cannot do that transformation.
But yeah, i understand what you mean. When you are coding and you get an error, sometimes it can be really hard to find the error, specially when it is a long code.
I always take a break when i am coding long files to check if there are any errors, and if there are none, i save them as a separate backup file, like that it will be easier to solve any mistake you encounter.
yeah I knew, I was just trying to confuse people
oh yeah
@ralph relatable, right