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A ve number greater than infinity? 
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#1
Jun2406, 12:32 AM

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. 


#2
Jun2406, 01:25 AM

P: 2,046

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 


#3
Jun2406, 02:33 AM

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#4
Jun2406, 02:48 AM

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A ve number greater than infinity?
Mainly, your answer doesn't work because 5 is a prime number.



#5
Jun2406, 07:03 AM

P: 24

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



#6
Jun2406, 07:07 AM

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Depends on your notion of smallness.



#7
Jun2506, 12:07 AM

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. 


#10
Jun2806, 11:53 PM

P: 194

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


#11
Jun2906, 07:21 AM

P: 925

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



#12
Jun2906, 07:51 AM

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#13
Jul306, 08:59 PM

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? 


#14
Jul406, 09:57 AM

P: 15,319

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.) 


#15
Jul606, 08:32 PM

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. 


#16
Jul606, 10:31 PM

P: 1,075




#17
Jul806, 12:12 PM

P: 26




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