Statement to prove:

Let x, y be in the reals with x < y.

If u is a real number with u > 0, show that there exists a rational number r such that x < ru < y.

We proved something similar to this in class. We proved the density of the rationals in the reals:

Claim: If x, y are real numbers and x < y, then there exists a rational number such that x < r < y.

Proof:

If x < 0 and y > 0 then x < 0 < y, and clearly 0 is rational.

If x ≥ 0: Since x < y, then 0 < y - x. By Archimedian Property (AP), there exists a natural number such that n(y - x) > 1 (useing "(y - x)" for "x" in AP and "1" for "y" in AP).

Note n(y - x) > 1 implies that ny > 1 + nx.

Let A = {k be a natural number: k > nx}. By AP, A is nonempty. By the Well Ordering Principle, A has a smallest element, say m in A is the smallest. Thus, m > nx ≥ m - 1 (m > nx since m is in A; Since m is the smallest, either (m - 1) is in N\A or (m - 1) = 0 if m - 1 and recall nx ≥ 0.

Note nx ≥ m – 1 implies 1 + nx ≥ m, but also m > nx so we have 1 + nx ≥ m > nx.

Since ny > 1 + nx, we have ny > m > nx which implies y > m/n > x and m/n is rational.

If x < 0 and y ≤ 0, apply the above to -x and -y.

Then yield r is rational with -y < r < -x which implies y > -r > x and -r is rational.

QED.

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So can I just basically state that u is a real number and u > 0 and work that into the proof?