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

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

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

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

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

since i²=-1,

then 0x∞=i²

hmm...

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

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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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,564

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

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

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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