1. Feb 2, 2012

cragar

1. The problem statement, all variables and given/known data
Given any 2 reals a<b there exists an irrational number t such that
a<t<b . It tells us to use a theorem that states there is a rational number between any 2 reals.
3. The attempt at a solution
so If we use this and pick reals of the form p=a-√2 and q=b-√2
so now we have a-√2<t<b-√2 , and t is a rational number. then we add √2
to everything so we get a<t+√2<b
so now t+√2 is irrational , so now we have an irrational between any 2 reals.

2. Feb 2, 2012

Dick

Again, no arguments with your reasoning. Do you have any doubts?

3. Feb 2, 2012

cragar

no, im just new to writing proofs, so I just wanted more experienced people to check my work.

4. Feb 2, 2012

deleted

5. Feb 3, 2012

checkitagain

From the highlighted above, you are stating that t is rational and that it is irrational.

6. Feb 3, 2012

cragar

t is rational , t+√2 is irrational

7. Feb 3, 2012

HallsofIvy

Yes, checkitagain's point was that you state the problem as "Given any 2 reals a<b there exists an irrational number t such that a<t<b" then later use the same letter, t, to represent a rational number.

8. Feb 3, 2012

cragar

ok so I just need to be careful with how I define my variables.

9. Feb 3, 2012

Shayes

check it again, checkitagain

10. Feb 3, 2012

Shayes

yes but honestly it looks fine to me

11. Feb 3, 2012

Shayes

just replace this

so now we have a-√2<t<b-√2 , and t is a rational number. then we add √2

with this

so now we have a-√2<s<b-√2 , and s is a rational number. then we add √2

and the other lines as necessary

12. Feb 5, 2012

checkitagain

There is no need to "check it again." That OP's post is as wrong now
because of the wrong variable as it was then.

Originally Posted by cragar:
------------------------------------------------------------
ok so I just need to be careful with how I define my variables.
------------------------------------------------------------

Stating "yes," and then it looks "fine" are contradictory.
That's why my post and HallsofIvy's accurate posts
are all that was needed.

@ cragar, once you have altered your proof with what you admitted
to in post # 8, then it will be "fine."

Last edited: Feb 5, 2012