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]

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.

 

[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]

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

That's funny.

 

[Rach3]

That's funny.

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]

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.

 

[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]

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.

 

[Tyris]

Here am a graph of 1/x.
http://www.mathsrevision.net/gcse/1overx.gif
5/x follows the same pattern.