# Thread: Intro to Analysis Proof help

1. ## Intro to Analysis Proof help

Hi, I'm self studying analysis, and I'm having problem with this proof. I've gotten most in my book so far but this has posed a problem.

I've attached the problem.

for part 1 I just said A is non empty since when x = 1/5 and n = 5 1 is an integer
for part 2 I let k = 1, x = 3/2 and showed that 1 < 3/2 < 2
For part 3 and 4 I have no idea what to do and i'm not sure whether i've done the prior correctly. I'd appreciate any help. Thanks!

2. ## Re: Intro to Analysis Proof help

Originally Posted by glambeth
Hi, I'm self studying analysis, and I'm having problem with this proof. I've gotten most in my book so far but this has posed a problem.

I've attached the problem.

for part 1 I just said A is non empty since when x = 1/5 and n = 5 1 is an integer
Then you have misunderstood the whole thing! The theorem you are trying to prove is that "If p is the not the square of an integer and $x^2= p$ then x is irrational". You cannot just choose a value for x. x is "given" by $x^2= p$. The crucial point here is "Assume that x is a positive rational number. What is the definition of "rational number"?

for part 2 I let k = 1, x = 3/2 and showed that 1 < 3/2 < 2
Again, you cannot assume a value for x. x is given by $x^2= p$ and you must use that x.

For part 3 and 4 I have no idea what to do and i'm not sure whether i've done the prior correctly. I'd appreciate any help. Thanks!

3. ## Re: Intro to Analysis Proof help

Originally Posted by glambeth
Hi, I'm self studying analysis, and I'm having problem with this proof. I've gotten most in my book so far but this has posed a problem.
I've attached the problem.
for part 1 I just said A is non empty since when x = 1/5 and n = 5 1 is an integer
for part 2 I let k = 1, x = 3/2 and showed that 1 < 3/2 < 2
For part 3 and 4 I have no idea what to do and i'm not sure whether i've done the prior correctly. I'd appreciate any help. Thanks!
Do not use examples in proofs.
Because $x\in\mathbb{Q}$ then there is an integer $n$ such that $nx\in J$ so $A\ne\emptyset.$

Use well-ordering to say that $A$ has a first term call it $m$.

Again $x$ cannot be integer so it is between two consecutive integers, $k or
$\\0.

You know that both $mx~\&~mk$ are integers (WHY?) thus $(mx-mk)x$ is also an integer.

But $(mx-mk)\notin A$ so there is your contradiction . (HOW & WHY??)

4. ## Re: Intro to Analysis Proof help

Wow! I misread that problem completely, I appreciate the help. I'll let you guys know if I have any problems with constructing my proof! Thanks again!

5. ## Re: Intro to Analysis Proof help

Could you explain how mx and mk are integers and why the last line is a contradiction?