There exists a real number such that an integer is greater than it, right?

Hello,

I'm confused about something.

-If there's a value "a" such that it's a real number, and there exists a "b" such that it's an integer and that b > a
-If there exists an "a" such that it's a real number and that, for all of "b" that's a real number, b> a

Both of these statements should be true, right? Because no matter what you can always add 1 more to any numbers. But apparently the first statement I listed is true, but the second is false.

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Originally Posted by Lord Darkin

Hello,

I'm confused about something.

-If there's a value "a" such that it's a real number, and there exists a "b" such that it's an integer and that b > a
-If there exists an "a" such that it's a real number and that, for all of "b" that's a real number, b> a

Both of these statements should be true, right? Because no matter what you can always add 1 more to any numbers. But apparently the first statement I listed is true, but the second is false.

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The first statement says there exist a "b" that is > than a.

The second statement says all b are > than a.

If our set is the reals, the first is true and the second is false.

Thanks!

Quote:

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If there's a value "a" such that it's a real number, and there exists a "b" such that it's an integer and that b > a

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I agree with dwsmith's post above, but, to be strict, the quoted statement is not well-formed from the standpoint of either mathematics or English. A sentence that has an "if" must also have a "then," even if it is only implied. The correct statement is, "If is a real number, then there exists an integer b such that b > a."