Maths can be weird....
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your example only shows that it goes to +-infinity. you never show that it is undefined.
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why cannot have two possible values of y for x=0?Originally posted by CoolMyth:Yep.
The idea is the same. You can't have two possible values for y for x = 0. The logic method and the calculus are similar in that sense.
draw the graph of x^2 + y^2=1
unit circle about the origin,
when x=0 y = +1 or -1 -
the fact that it goes to 2 different values shows that its undefined lahOriginally posted by Gasfene:your example only shows that it goes to +-infinity. you never show that it is undefined. -
again no.Originally posted by hisoka:the fact that it goes to 2 different values shows that its undefined lah
For eg, draw the graph of
y= -1 for x<=0
= 1 for x>=0
notice that when x=0, y = 1 or -1. It goes to 2 different values but x is defined here. This graph is just discontinuous. -
I'll refer to your own proof.Originally posted by Gasfene:why cannot have two possible values of y for x=0?
draw the graph of x^2 + y^2=1
unit circle about the origin,
when x=0 y = +1 or -1
You said let 1/0 = x --> x*0 = 1
But then x*0 = 0 rite?
So in that similar manner, you can't have two possible values of y = x*0 when x = 0. That's what I'm trying to say.
In maths, as long as you can produce a counterexample for any claim you made to be true, the claim is invalid. No matter how many "successful" examples I can give, as long as there is one counterexample for which the claim is false, the entire thing is wrong.
You can give the logic proof surely you know this principle right?
In that sense, the calculus proof shows the counterexample where the right limit is not equal to the left limit thus showing the limit at that point does not exist and thus the point x = 1/0 is undefined.
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i'm starting not to understand what's in this thread.
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That's why maths is weird.Originally posted by finoq:i'm starting not to understand what's in this thread. -
agreedOriginally posted by CoolMyth:That's why maths is weird. -
For eg, draw the graph ofOriginally posted by CoolMyth:I'll refer to your own proof.
You said let 1/0 = x --> x*0 = 1
But then x*0 = 0 rite?
So in that similar manner, you can't have two possible values of y = x*0 when x = 0. That's what I'm trying to say.
In maths, as long as you can produce a counterexample for any claim you made to be true, the claim is invalid. No matter how many "successful" examples I can give, as long as there is one counterexample for which the claim is false, the entire thing is wrong.
You can give the logic proof surely you know this principle right?
In that sense, the calculus proof shows the counterexample where the right limit is not equal to the left limit thus showing the limit at that point does not exist and thus the point x = 1/0 is defined.
y= -1 for x<=0
= 1 for x>=0
notice that when x=0, y = 1 or -1. It goes to 2 different values but x is defined here. the limit when x->0- = -1
the limit when x->0+ = 1
In this case the limit is different and hence the limit does not exist. But at f(0) is well defined. So you people only prove limit does not exist but not why it is undefined. -
Excuse me, we're talking purely about the question of whether 1/0 is defined right?Originally posted by Gasfene:For eg, draw the graph of
y= -1 for x<=0
= 1 for x>=0
notice that when x=0, y = 1 or -1. It goes to 2 different values but x is defined here. the limit when x->0- = -1
the limit when x->0+ = 1
In this case the limit is different and hence the limit does not exist. But at f(0) is well defined. So you people only prove limit does not exist but not why it is undefined.
So, the question is asking whether y is defined at x = 0 where y = 1/x, correct?
For other polynomials equations, y = f(x). y can defined at x = 0 for all I care. In fact I can list a lot of them.
Let's get back to question, shall we?
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you use y=1/x but you only show that the limit does not exist at x=0.Originally posted by CoolMyth:Excuse me, we're talking purely about the question of whether 1/0 is defined right?
So, the question is asking whether y is defined at x = 0 where y = 1/x, correct?
For other polynomials equations, y = f(x). y can defined at x = 0 for all I care. In fact I can list a lot of them.
Let's get back to question, shall we?
You never show that it is undefined there so your proof is 'out of topic' here. -
hmm
I have two maths questions
1) if 1/-0.000manyzeros00001 is a large -ve number and 1/0.000000manyzeros000001 is a large +ve number.. does 1/0 tends to positive or negative infinity??
2) What is 0/0?? can 0/0 be 1 since there is 1 part of "nothing" in 1 part of "nothing"???
I hate taylor's, euler's whatever's expansions too
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i said it goes to not it is lah...............Originally posted by Gasfene:again no.
For eg, draw the graph of
y= -1 for x<=0
= 1 for x>=0
notice that when x=0, y = 1 or -1. It goes to 2 different values but x is defined here. This graph is just discontinuous. -
1) limit does not exist. Nothing has been proven whether it is undefined or not.Originally posted by DriftingGuy:hmm
I have two maths questions
1) if 1/-0.000manyzeros00001 is a large -ve number and 1/0.000000manyzeros000001 is a large +ve number.. does 1/0 tends to positive or negative infinity??
2) What is 0/0?? can 0/0 be 1 since there is 1 part of "nothing" in 1 part of "nothing"???
I hate taylor's, euler's whatever's expansions too -
my point of showing that graph is to tell you even if approach from two different directions, you get two different limits thus limit not defined but the real function can still be defined at 0Originally posted by hisoka:i said it goes to not it is lah............... -
ooi dun make me say that your prove wrong cos you never prove that 0 time x is zero are that 1/somethign exist horOriginally posted by Gasfene:1) limit does not exist. Nothing has been proven whether it is undefined or not. -
huh??? your example does not show that btw you need to rethink. unless you means your later postsOriginally posted by Gasfene:my point of showing that graph is to tell you even if approach from two different directions, you get two different limits thus limit not defined but the real function can still be defined at 0 -
we assume x is some real number remember?Originally posted by hisoka:ooi dun make me say that your prove wrong cos you never prove that 0 time x is zero are that 1/somethign exist hor
so any real number times 0 = 0 (its a known fact) so contradiction. -
You guys need to read more of logic, basics of maths or get a clear head before commenting further. Thanks shall not reply further. Get a maths pro to advice if you insist your answer is right or what.
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pls proof that 0 = 0 is valid.Originally posted by Gasfene:we assume x is some real number remember?
so any real number times 0 = 0 (its a known fact) so contradiction.
and prove that 0 is a real number. please proof that the definition of real number is correct.
btw known fact is not equal to correct by your definitions remember?? -
nay you haven't proven your logic is right anyway.Originally posted by Gasfene:You guys need to read more of logic, basics of maths or get a clear head before commenting further. Thanks shall not reply further. Get a maths pro to advice if you insist your answer is right or what. -
sighs get a maths pro to advice i say again.
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yeah pls get a maths pro to advice yourself pls. thxOriginally posted by Gasfene:sighs get a maths pro to advice i say again.
haiz......... -
0=0 can be proven but it is very very abstract and it is not easily understandable by anyone. I need to dig my notes first,Originally posted by hisoka:pls proof that 0 = 0 is valid.
and prove that 0 is a real number. please proof that the definition of real number is correct.
btw known fact is not equal to correct by your definitions remember??
Definition means it is defined to be so. There's isnt a need to prove it. Its the same as C is defined as Carbon. like that you prove to me C is Carbon? -
go dig then say bah. btw i will ask you to prove every thing you do hor. including all x+ y and x/y and what not.Originally posted by Gasfene:0=0 can be proven but it is very very abstract and it is not easily understandable by anyone. I need to dig my notes first,
Definition means it is defined to be so. There's isnt a need to prove it. Its the same as C is defined as Carbon. like that you prove to me C is Carbon?
it'll be fun to watch if nothign else
why no need to prove that definition no need to prove? pls proof that.