Proof about irrationals between reals.

[cragar]

Homework Statement

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.

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.

 

[Dick]

Homework Statement

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.

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.

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

 

[cragar]

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

 

[jbunniii]

deleted

 

[checkitagain]

Homework Statement

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.

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.

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

These contradict each other, yes?

 

[cragar]

t is rational , t+√2 is irrational

 

[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.

 

[cragar]

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

 

[Shayes]

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

These contradict each other, yes?

check it again, checkitagain

 

[Shayes]

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

yes but honestly it looks fine to me

 

[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

 

[checkitagain]

check it again, 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.
------------------------------------------------------------

yes but honestly it looks fine to me

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."