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Confusion over 'there exists' in a proof!

So I have to prove the following statement. If x is a rational number then (this implies) there exists a natural number n such that nx is an integer.

Does this mean that I have to prove/disprove that there is a single number n, which when multiplied by any rational gives an integer OR is it that I have to prove/disprove for every rational number x, there is a n value which can make there product an integer?

The 'there exists' is confusing me :s-smilie:

Thanks for any help!
Reply 1
Avatar for davros
davros
16
Original post by Student10011
So I have to prove the following statement. If x is a rational number then (this implies) there exists a natural number n such that nx is an integer.

Does this mean that I have to prove/disprove that there is a single number n, which when multiplied by any rational gives an integer OR is it that I have to prove/disprove for every rational number x, there is a n value which can make there product an integer?

The 'there exists' is confusing me :s-smilie:

Thanks for any help!


The second interpretation is the correct one.

The first one can't possibly be true - there's no such integer n which can contain all the possible divisors of any rational number :smile:
Reply 2
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Student10011
OP
Original post by davros
The second interpretation is the correct one.

The first one can't possibly be true - there's no such integer n which can contain all the possible divisors of any rational number :smile:



Thanks! I knew which one was true, just unsure which interpretation of 'there exists' was correct. Thanks for clearing it up!:wink:

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