A_B
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Hi,
In Rudin's Principles of Mathematical Analysis there's the following proof that Q is dense in R.
Theorem: If x,y\in \mathbb{R} and x < y there exists a p \in \mathbb{Q} such that x<p<y.
Proof: Since x<y, we have y-x>0. It follow from the Archimedian property that there is a positive integer n such that
n(y-x)>1.
We again apply the Archimedian property to find positive integers m_1 a,d m_2 such that m_1>nx and m_2>-nx. Then
-m_2<nx<m_1.
Hence there is an integer m (with -m_2\leq m\leq m_1) such that
m-1\leq nx < m.
We combine the inequalities to get
nx < m \leq 1+nx < ny.
n is positive so
x < \frac{m}{n} < y.
Which proves that \mathbb{Q} is dense in \mathbb{R}.
How one concludes that the m in the red bit exists is what's troubling me.
Thanks
In Rudin's Principles of Mathematical Analysis there's the following proof that Q is dense in R.
Theorem: If x,y\in \mathbb{R} and x < y there exists a p \in \mathbb{Q} such that x<p<y.
Proof: Since x<y, we have y-x>0. It follow from the Archimedian property that there is a positive integer n such that
n(y-x)>1.
We again apply the Archimedian property to find positive integers m_1 a,d m_2 such that m_1>nx and m_2>-nx. Then
-m_2<nx<m_1.
Hence there is an integer m (with -m_2\leq m\leq m_1) such that
m-1\leq nx < m.
We combine the inequalities to get
nx < m \leq 1+nx < ny.
n is positive so
x < \frac{m}{n} < y.
Which proves that \mathbb{Q} is dense in \mathbb{R}.
How one concludes that the m in the red bit exists is what's troubling me.
Thanks