Is There Always a Real Number Between Two Arbitrary Real Numbers?

AI Thread Summary
For any two distinct real numbers x and y, there exists at least one real number z such that x < z < y, which can be demonstrated using the Archimedean Property. The midpoint (x+y)/2 serves as a valid example of such a z. The discussion clarifies that the original statement is true only if x and y are distinct, and that the property also holds for rational and complex numbers. The necessity of the real number condition is debated, emphasizing the broader applicability of the concept. Overall, the existence of a real number between two distinct arbitrary real numbers is affirmed.
courtrigrad
1,236
2
Hello all

I need help with the following proofs

1. If x and y are arbitrary real numbers, prove that there is at least one real z satisfying x < z < y. (Do I just use the Archimedian Property?)

Thanks
 
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courtrigrad said:
(Do I just use the Archimedian Property?)

Yep. :smile:
 
No. and it isn't true as stated.

x and y must be distinct, and the requirement of being R is unnecessary, since it is true for Q (and C).


what is (x+y)/2?
 
Good catch. I was thinking of rational numbers between arbitrary reals.
 
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