- #1
completenoob
- 26
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Hello,
I have decided to study analysis on my own and am starting with principles of mathematical analysis by rudin.
I am having trouble understanding pg. 9 on the density of Q in R, part b.
It states:
If [itex]x \in R, y \in R [/itex] and [itex] x<y [/itex] the there exists a [itex] p \in Q [/itex] such that [itex]x < p < y[/itex]
Proof:
Since [itex] x<y[/itex], we have [itex] y-x>0[/itex] and the Archemedian Property furnishes a positive integer n such that:
[itex]n(y-x)>1[/itex]
Applying the AP again, to obtain positive integers m1 & m2 such that [itex]m1>nx[/itex], [itex]m2>-nx[/itex]
Then: [itex]-m2<nx<m1 [/itex]
Hence there is an integer m such that [itex]m-1 \le nx<m [/itex]...
Can someone explain to me the last line? Where does this less then or equal come from?
I have decided to study analysis on my own and am starting with principles of mathematical analysis by rudin.
I am having trouble understanding pg. 9 on the density of Q in R, part b.
It states:
If [itex]x \in R, y \in R [/itex] and [itex] x<y [/itex] the there exists a [itex] p \in Q [/itex] such that [itex]x < p < y[/itex]
Proof:
Since [itex] x<y[/itex], we have [itex] y-x>0[/itex] and the Archemedian Property furnishes a positive integer n such that:
[itex]n(y-x)>1[/itex]
Applying the AP again, to obtain positive integers m1 & m2 such that [itex]m1>nx[/itex], [itex]m2>-nx[/itex]
Then: [itex]-m2<nx<m1 [/itex]
Hence there is an integer m such that [itex]m-1 \le nx<m [/itex]...
Can someone explain to me the last line? Where does this less then or equal come from?