# a -ve number greater than infinity?

by AlbertEinstein
Tags: greater, infinity, number
 P: 113 Please follow the following arguments. 5/3=1.66 5/2=2.5 5/1=5 5/0.5=10 ..... .... 5/0=infinity and then 5/(-1)= -5 What you see? As the denominator is decreased the right hand side answer increases. The denominator becomes 3 then 2,then 1, then 0,then -1 ; and the answer increases, therefore -5 must be grater than infinity. Where's the flaw. Please illustrate.
 P: 2,048 The curve 5/x is an example of a rectangular hyperbola. It's an asymptotic curve. http://en.wikipedia.org/wiki/Asymptote http://mathworld.wolfram.com/Asymptote.html
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 Quote by AlbertEinstein and the answer increases, therefore -5 must be grater than infinity. Where's the flaw. Please illustrate.
In assuming that because for some set of values f(x) is increasing it is always increasing.

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## a -ve number greater than infinity?

Mainly, your answer doesn't work because 5 is a prime number.
 P: 24 As one divides by smaller and smaller numbers, the output figure approaches but does not reach infinity, no?
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 Depends on your notion of smallness.
 P: 38 Please follow the following arguments. 5=5 4=4 3=3 2=2 1=1 0=0 -1=-1 What do you see? As the left hand side decreases, the right hand side is greater than or equal to zero. Therefore, -1 must be greater than or equal to zero. Ja? Just because a property holds true for a certain range of numbers, it doesn't mean the pattern will follow for numbers outside that range. That's the flaw.
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 Quote by AKG Mainly, your answer doesn't work because 5 is a prime number.
That's funny.
P: 318
 Quote by Gokul43201 That's funny.
It is?
 P: 194 Ya, its funny because AKG's answer isn't right. ja ja
 P: 927 5/0 does not equal infinity. The limit of 5/x as x approaches 0 equals inifinty.
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 Quote by daveb 5/0 does not equal infinity. The limit of 5/x as x approaches 0 equals inifinty.
In absolute value, yes. But else you have to mind the sign, depending on whether you're approaching 0 from the left or right, you get -inf resp. +inf.
 P: 5 I don't see how you can say that -1 is greater than or equal to 0. I might be horribly wrong but what I think you're doing is just equating numbers & two equal numbers are always equal .. under no circumstances can be greater than or equal to. Look at this order: 1=1 9=9 8=8 -1=-1 Don't you see?
 P: 15,294 Albert: Please follow the following arguments. (x^2=y) 3^2=9 2^2=4 1^2=1 0^2=0 -1^2=1 What you see? As the x value is decreased, the y value (right hand side answer) decreases. The x value becomes 3 then 2,then 1, then 0,then -1 ; and the answer (y) decreases, therefore 1 must be less than zero. Now where's the flaw? (Hint: the only flaw is the conclusion that a given function must result in a straight and continuous line.)
 P: 218 Okay...let me try to hit this topic. Graphing 5/x you will have a line that going from negative infinity to positive infinity it ALWAYS DECREASES. it never increases. However, it starts out negative and ends up positive. Lesson learned today: Don't mess with the division by zero.
P: 1,076
 Quote by Robokapp Okay...let me try to hit this topic. Graphing 5/x you will have a line that going from negative infinity to positive infinity it ALWAYS DECREASES. it never increases. However, it starts out negative and ends up positive. Lesson learned today: Don't mess with the division by zero.
No you do not have a line. You have a hyperbola in the first and third quandrants of the cartesian plane.
 P: 26 Here am a graph of 1/x. clicky 5/x follows the same pattern.

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