Zero x infinity = -1 ?

  • Thread starter bs
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  • #1
bs
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Zero x infinity = -1 ???!!!

when two lines meet perpendicularly, we hv the formula v
m1 x m2 = -1

if a horinzontal line meet v a vertical line,by using same formula ,we hv
m1 x m2 = -1
zero x infinity = -1

so..does it make sense??!! :eek:
 

Answers and Replies

  • #2
https://www.physicsforums.com/showthread.php?t=80945
 
  • #3
HallsofIvy
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bs said:
when two lines meet perpendicularly, we hv the formula v
m1 x m2 = -1

if a horinzontal line meet v a vertical line,by using same formula ,we hv
m1 x m2 = -1
zero x infinity = -1

so..does it make sense??!! :eek:
No, "infinity" is not a real number. Saying that a vertical line has slope "infinity" is just a way of saying that a vertical line does not have a slope. (In fact, it would make just as much sense to say that a vertical line has slope "negative infinity".) There are a variety of numbers systems that include some notion of "infinity" but in those multiplication of a number by infinity is either not defined or gives something other than -1. I don't know of any number system that defines multiplication of zero by infinity to give -1.
 
  • #4
Alkatran
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Multiplying 0 by infinity is like dividing 0 by 0, inderterminate. You can get any answer you want. For example:

1:
[tex]
\lim_{x\rightarrow\infty} x * \frac{1}{x^2} = 0 * \infty
[/tex]
[tex]
\lim_{x\rightarrow\infty} x * \frac{1}{x^2} = 0
[/tex]

[tex]
0 * \infty = 0
[/tex]

2:
[tex]
\lim_{x\rightarrow\infty} x * \frac{1}{x} = 0 * \infty
[/tex]
[tex]
\lim_{x\rightarrow\infty} x * \frac{1}{x} = 1
[/tex]

[tex]
0 * \infty = 1
[/tex]

3:
[tex]
\lim_{x\rightarrow\infty} 2 * x * \frac{1}{x} = 0 * \infty
[/tex]
[tex]
\lim_{x\rightarrow\infty} 2 * x * \frac{1}{x} = 2
[/tex]

[tex]
0 * \infty = 2
[/tex]
 
  • #5
pallidin
2,209
2
Alkatran said:
Multiplying 0 by infinity is like dividing 0 by 0, inderterminate. You can get any answer you want. For example:

1:
[tex]
\lim_{x\rightarrow\infty} x * \frac{1}{x^2} = 0 * \infty
[/tex]
[tex]
\lim_{x\rightarrow\infty} x * \frac{1}{x^2} = 0
[/tex]

[tex]
0 * \infty = 0
[/tex]

2:
[tex]
\lim_{x\rightarrow\infty} x * \frac{1}{x} = 0 * \infty
[/tex]
[tex]
\lim_{x\rightarrow\infty} x * \frac{1}{x} = 1
[/tex]

[tex]
0 * \infty = 1
[/tex]

3:
[tex]
\lim_{x\rightarrow\infty} 2 * x * \frac{1}{x} = 0 * \infty
[/tex]
[tex]
\lim_{x\rightarrow\infty} 2 * x * \frac{1}{x} = 2
[/tex]

[tex]
0 * \infty = 2
[/tex]


Amazing. Off the subject, but a comment nonetheless: Perhaps your examples demonstrates how "all things are possible"
 
  • #6
Alkatran
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pallidin said:
Amazing. Off the subject, but a comment nonetheless: Perhaps your examples demonstrates how "all things are possible"

No, it demonstrates how you can't multiply 0 by infinity and get an answer because its indeterminate (except for special cases with limits...)
 
  • #7
toocool_sashi
24
0
if u read the theorem on m1*m2 = -1 carefully...u shud notice 1 thing....it is SPECIFICALLY SAID..(in my notes @least)....m1 m2 are slopes of lines NOT PARALLEL to either of the axes. in addition...by DEFINITION...the axes themselves are mutually perpendicular(i have come across problems wherin the axes are not perpendicular too....i remember doing a problem 2 find out distance between 2 pts when axes are inclined @ 30degrees). If u proceed with the proof for m1*m2=-1...u will find an error (dividing by zero or multiplying by infinity) in the penultimate step..check it out ;)
 

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