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#1 2005-08-15 22:40:04

wcy
Member
Registered: 2005-08-04
Posts: 117

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 ??

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#2 2005-08-15 22:56:45

wcy
Member
Registered: 2005-08-04
Posts: 117

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 (2005-08-15 22:57:00)

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#3 2005-08-15 23:16:41

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

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.

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#4 2005-08-16 19:26:52

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,842

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.

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#5 2005-08-22 21:00:47

wcy
Member
Registered: 2005-08-04
Posts: 117

Re: 0x∞=-1 ???

since i²=-1,
then 0x∞=i²
hmm...

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#6 2005-09-05 11:04:53

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

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.

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#7 2005-09-05 11:55:46

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

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 (2005-09-05 11:56:36)


igloo myrtilles fourmis

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#8 2005-10-30 13:17:36

God
Member
Registered: 2005-08-25
Posts: 59

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 (2005-10-30 13:18:02)

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#9 2005-10-30 17:18:30

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,552

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

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