He's just mad because he got held back in school for failure to understand basic concepts of arithmetic.
Can you prove 1 = 2?
- Thread starter secondeye
- Start date
I don't think you guys get what I am saying.
When you have an algebraic equation it is supposed to be representative of all numbers, therefore it is a valid equation. However if this does not work for one instance (the one described) then it is proved not to work overall.
An example would be this equation:
x=(a-b)/(a-b)
where in ALMOST all cases x=1 and the equation is correct
BUT
if a=b then the equation is incorrect and does not work as x cannot be described.
but as an algebraic expression it is perfectly correct. How? Does this mean that for all algebraic formula to be correct you need to add in real numbers to prove it? If so then doesn't this show a flaw in algebra?
When you have an algebraic equation it is supposed to be representative of all numbers, therefore it is a valid equation. However if this does not work for one instance (the one described) then it is proved not to work overall.
An example would be this equation:
x=(a-b)/(a-b)
where in ALMOST all cases x=1 and the equation is correct
BUT
if a=b then the equation is incorrect and does not work as x cannot be described.
but as an algebraic expression it is perfectly correct. How? Does this mean that for all algebraic formula to be correct you need to add in real numbers to prove it? If so then doesn't this show a flaw in algebra?
Actually, the correct algebraic expression would state a =/= b as a limit. Most of the time this isn't stated because the a = b answer is trivial.
But still, the equation you gave below works perfectly if a = b, it's telling you the answer- it's just that the answer isn't defined, so how can you expect it to give you a definite one, how is that a flaw of algebra?
Sketch the plot of the equation and you'll see what you're asking for.
(BTW, If you haven't started using limits in mathematics, I'm also guessing you probably haven't had much education in the subject. It's standard above A-level (UK), High School/Freshman College (US) level.)
But still, the equation you gave below works perfectly if a = b, it's telling you the answer- it's just that the answer isn't defined, so how can you expect it to give you a definite one, how is that a flaw of algebra?
Sketch the plot of the equation and you'll see what you're asking for.
(BTW, If you haven't started using limits in mathematics, I'm also guessing you probably haven't had much education in the subject. It's standard above A-level (UK), High School/Freshman College (US) level.)
Correction, the a=b isn't strictly speaking a trivial solution, just not a useful one.
WTF that's random, I graduated IC in Physics.
But to the point, what exactly is incorrect about the equation? It's telling you precisely what you've asked of it.
Edit: Ok, reread your post above and I see what you're getting at. But it's hardly a refutation of algebra, but just an acknowledgment that not all numbers that exist are real.
It seems to me that you're repeating what your teacher told you and passing it off as your own opinion.
And it's still stupid, GTFO.
Here, let me try. Don't think it would make any sense but no harm to give it a try.
A simple formula:
Let 1=a and 2=b
Now,
0 = 0
a x 0 = b x 0
Canceling 0 as it is present in both sides.
Thus, a = b or 1 = 2 ...
Have I do it correctly? :hitit_sml:
A simple formula:
Let 1=a and 2=b
Now,
0 = 0
a x 0 = b x 0
Canceling 0 as it is present in both sides.
Thus, a = b or 1 = 2 ...
Have I do it correctly? :hitit_sml:
I give up. You're absolutely right. Algebra is fatally flawed because of what you posted above. In fact, I think I will stop using algebra because of this.
"Cancelling 0" is the same as dividing by zero. You can't do that.
By using calculus?
As a matter of fact, most of the confusion in this thread seems to be the result of people not understanding that algebra is just a small subset of math with limited application.
i'd say algebra has ubiquitous application.
By using calculus?
As a matter of fact, most of the confusion in this thread seems to be the result of people not understanding that algebra is just a small subset of math with limited application.
you realize calculus relies on the existence of algebra, correct?