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#1 2005-08-16 20:40:04
- wcy
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0x∞=-1 ???
Under the topic of linear graphs, we are taught that if two straight lines are perpendicular to each other, then their product of gradients is equal to -1.
Since a line with gradient 0 (horizontal line)is perpendicular to a line with gradient ∞ (vertical line), isn't 0x∞=-1 ??
#2 2005-08-16 20:56:45
- wcy
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Re: 0x∞=-1 ???
Proof of following "theorem" using vectors method.
Let the first line be a vector (x,y) (vertical matrix)
Let the 2nd line be a perpendicular vector (p,q)
cos 90=0=(x,y).(p,q)=xp+yq (dot product)
yq=-xp
Therefore, gradient1xgradient2=(y/x)(q/p)=yq/xp=-xp/xp=-1
Last edited by wcy (2005-08-16 20:57:00)
#3 2005-08-16 21:16:41
- mathsyperson
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Re: 0x∞=-1 ???
That is dependant on the fact that n/0=∞, which I don't like. I prefer the idea that as 6x0=0, then 0/0=6, with 6 being replaced with any number. As 0/0 can take any value, then it is undefined.
Then again, that would mean that 0xn=-1, which means that -1/n=0, but n can take any value and if we give it the value of 1 then -1/1=-1, so we have a contradiction. I think the moral is that anything involving 0 is too strange to try to deal with.
Why did the vector cross the road?
It wanted to be normal.
#4 2005-08-17 17:26:52
- ganesh
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Re: 0x∞=-1 ???
Too complex for the human brain to comprehend.....
wcy showed us 0 x ∞ = -1
and there seems to be no loophole in the proof.
I was almost convinced that 0 x ∞ = 0 x 1/0 = 1, now I'd have to tell my mind it is ±1.
When we take the value -1,
0 x ∞ = -1
Taking square root on both sides,
√ (0 x ∞) = √-1
0 x √ ∞ = i
Therefore, √ ∞ = -i/0 !
wcy, your theory appears to be right!
Character is who you are when no one is looking.
#6 2005-09-06 09:04:53
- Zhylliolom
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Re: 0x∞=-1 ???
ganesh wrote:
I was almost convinced that 0 x ∞ = 0 x 1/0 = 1, now I'd have to tell my mind it is ±1.
With "∞ = 1/0" reasoning, we could argue that 2/0 is also ∞, and so on. From this, 0 x ∞ could be any number. Letting n/0 = ∞ is just illegal to me. True, n/m gets increasingly closer to ∞ as m gets closer to 0(from the right), but when m actually is 0, we don't have any sensible number to assign to it. Assigning ∞ to n/0 doesn't make any sense unless n = 0, making it so 0 x ∞ = 0, but then we run into the classic indeterminate form 0/0, which can be any number. As mathsyperson said, dealing with 0 like this brings us into an area of mathematics which we cannot explain.
#7 2005-09-06 09:55:46
- John E. Franklin
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Re: 0x∞=-1 ???
But the slope of vertical line can also be negative infinity too!
And the slope of a horizontal line could be negative zero!!
Last edited by John E. Franklin (2005-09-06 09:56:36)
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#8 2005-10-31 12:17:36
- God
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Re: 0x∞=-1 ???
General comments:
∞ is not a slope. The gradient of a vertical line is not ∞, since one could argue that it is also -∞, which is completely different.
0 x ∞ = -1
2 x 0 x ∞ = -1 x 2
0 x ∞ = -2
-1 = -2
You're right, this is way to complicated for the human mind to comprehend... meep
I'm going to stick to my safe and solid viewpoint that infinity is not a number until someone enlightens me.
Last edited by God (2005-10-31 12:18:02)
#9 2005-10-31 16:18:30
- MathsIsFun
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Re: 0x∞=-1 ???
Well, if God says it isn't a number, who are we to argue?
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman