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Real numbers

Prove that there is a rational number in the open interval (a,b),b>a, (a,b), b>a, where a,b∈R a, b \in \mathbb{R} .
(edited 7 years ago)
http://homepages.warwick.ac.uk/~masdbl/w5.pdf

Look at Assignment 1. See if that helps you.
Original post by Ano123
Prove that there is a rational number in the open interval (a,b),b>a, (a,b), b>a, where a,b∈R a, b \in \mathbb{R} .


a+b2\frac{a+b}{2} is rational and within the range for any a,b∈Ra,b \in \mathbb{R} thus statement is true. Not sure how to formally structure this sort of proof.
(edited 7 years ago)
Original post by RDKGames
a+b2\frac{a+b}{2} is rational and within the range for any a,b∈Ra,b \in \mathbb{R} thus statement is true. Not sure how to formally structure this sort of proof.


Let a=2a=\sqrt{2} and b=1+2b=1+\sqrt{2}

a+b2\frac{a+b}{2} is not rational.
(edited 7 years ago)
Original post by Math12345
Let a=2a=\sqrt{2} and b=1+2b=1+\sqrt{2}

(a+b)/2 is not rational.


oh lol missed that where they don't say a and b are rationals, would be easy otherwise.
(edited 7 years ago)
Reply 5
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Zacken
22
What have you tried?
Reply 6
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B_9710
18
If b−a<1 b-a<1 , then b−a=k,k∈R+,k<1 b-a=k, k\in \mathbb{R}^+ , k<1, if we multiply by some real constant n n so that n(b−a)≥1, n(b-a)\geq 1, (nk≥1 nk\geq 1 ) then we can say for sure that some integer m m lies in that interval which is of course rational. So from this we can deduce that na<m<nb na<m<nb , dividing by n n we get a<m/n<b a<m/n<b .
If b−a>1 b-a>1 then an integer lies between the a and b which is obviously rational. This completes the proof.
(edited 7 years ago)

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