a -ve number greater than infinity?

AlbertEinstein

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.

neutrino

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

matt grime

In assuming that because for some set of values f(x) is increasing it is always increasing.

AKG

Mainly, your answer doesn't work because 5 is a prime number.

Crusty

As one divides by smaller and smaller numbers, the output figure approaches but does not reach infinity, no?

arildno

Depends on your notion of smallness.

gnomedt

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.

Gokul43201

That's funny.

Rach3

It is?

TriTertButoxy

Ya, its funny because AKG's answer isn't right.
ja ja

daveb

5/0 does not equal infinity. The limit of 5/x as x approaches 0 equals inifinty.

TD

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.

Jahnabi

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?

DaveC426913

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.)

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.

d_leet

No you do not have a line. You have a hyperbola in the first and third quandrants of the cartesian plane.

Tyris

Here am a graph of 1/x.
clicky
5/x follows the same pattern.