Understanding the Behaviour of 1/a: Even or Odd?

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even or odd?

if a is an even number , what can we say about 1/a ;
similarly if a is an odd number , what can we say about 1/a?
 
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Numbers can only be even or odd if they're integers. Is 1/a going to be an integer?
 


Pretty much nothing in either case. When we talk about even or odd numbers, we're really talking about even or odd integers; that is, whether the remainder is 0 or 1 when we divide by 2. The reciprocals of integers aren't generally integers.
 


You can say...that if a is a random odd number, that it is more likely that a is larger than if a is a random even number.

1 3 5 7 9 = 25
2 4 6 8 10 = 30

I think I got that right *blinks*.
 


G037H3 said:
You can say...that if a is a random odd number, that it is more likely that a is larger than if a is a random even number.

No.

First, there's no such thing as a uniform random distribution on the integers. Second, if you take the limit of the uniform distribution on [0, N] as N increases without bound, you get probability 1/2 for both.
 


if we cannot decide weather a fractional number is even or odd then how can we tell that weather a negative number raised to a fractional power is positive or negative?

given answer is to be a real number...
 


If a negative number, raised to a fractional power, is real, then that power is defined to be positive. That is, we define a^{1/2} to be the positive number, x, such that x^2= a.
 


If x is a negative number, x1/n is:
Not a real number if n is even. Any real number raised to an even power gives a positive number (or 0), so x1/n doesn't exist.

Negative if n is an odd number. A positive number raised to an odd power gives another positive number, so x1/n is always negative.

But this has nothing to do with deciding whether 1/n is even or odd, it's just properties of real numbers
 


CRGreathouse said:
No.

First, there's no such thing as a uniform random distribution on the integers. Second, if you take the limit of the uniform distribution on [0, N] as N increases without bound, you get probability 1/2 for both.

It *tends* to 1/2. It isn't 1/2 to begin with. ._.
 
  • #10


Can't you say that if a is even 1/a will terminate, and if a is odd 1/a will be infinite repeating? omitting when a = 1.
 
  • #11


Diffy said:
Can't you say that if a is even 1/a will terminate, and if a is odd 1/a will be infinite repeating? omitting when a = 1.
No, that's not true. 6 is even, but 1/6 has an infinitely repeating decimal representation - .1666666...
 
  • #12


G037H3 said:
It *tends* to 1/2. It isn't 1/2 to begin with. ._.

The probabilities *tend* to 1/2, but the limit of them *is equal to* 1/2. Of course this limit need not be the probablity of anything.
 
  • #13


willem2 said:
The probabilities *tend* to 1/2, but the limit of them *is equal to* 1/2. Of course this limit need not be the probablity of anything.

If using every even and every odd number, then yes. But with a real application, the numbers are usually obviously finite. >.>
 
  • #14


G037H3 said:
It *tends* to 1/2. It isn't 1/2 to begin with. ._.

That's incorrect.
 
  • #15


G037H3 said:
If using every even and every odd number, then yes. But with a real application, the numbers are usually obviously finite. >.>

You'll have to say how you sample them. Depending on that answer you might find odds larger on average, evens larger on average, or neither larger on average.
 
  • #16


CRGreathouse said:
You'll have to say how you sample them. Depending on that answer you might find odds larger on average, evens larger on average, or neither larger on average.

as a range with an equal number of even and odd integers
 
  • #17


G037H3 said:
as a range with an equal number of even and odd integers

That's pretty arbitrary, why not one more odd than even? Is the practical situation as apparent when it comes to testing different things? If you pick a random positive integer, what is the probability for that it starts with the digit 1? Given that you can pick at random this probability must exist. To what would a "practical situation" determine this value?

You may work it out if you want, but you'll realize that if you start out with the numbers [1,N] and calculating the probability for this collection, your probability will not converge as N grows towards infinity. Hence two "practical situations" can give two completely different answers, and you can never choose N so large that the difference will be negligible.
 
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