Proof: there is M s.t for all r>M, 1/2r < 1/100

[Aziza]

"Prove that there exists an odd integer M such that for all real numbers r larger than
M, $\frac{1}{2r} < \frac{1}{100}$"

How to express this using logical symbols?

Literally I understand this statement as:
$(\exists M\in$[PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png$)[($M is odd$)\wedge(\forall r\inℝ)(r>M\wedge\frac{1}{2r} < \frac{1}{100})]$

But all r in ℝ cannot be greater than some M. There will always be r<=M if the only restriction on r is that it is in ℝ. So this statement can't be proven true. But would it be wrong to interpret the statement as follows:

$(\exists M\in$[PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png$)[($M is odd$)\wedge(\forall r\inℝ)(r>M\Rightarrow\frac{1}{2r} < \frac{1}{100})]$

(the last "and" changed to "imples")...then M=51 would prove the statement true...

 

[Simon Bridge]

"Prove that there exists an odd integer M such that for all real numbers r larger than
M, $\frac{1}{2r} < \frac{1}{100}$"

How to express this using logical symbols?

Literally I understand this statement as:
$(\exists M\in$[PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png$)[($M is odd$)\wedge(\forall r\inℝ)(r>M\wedge\frac{1}{2r} < \frac{1}{100})]$

$$\forall r \in \mathbb{R} > M \; ; \; \exists M = 2n-1 \; \wedge \; n \in \mathbb{Z} \; : \; \frac{1}{2r} < \frac{1}{100}$$

But all r in ℝ cannot be greater than some M. There will always be r<=M if the only restriction on r is that it is in ℝ.

but that's not the only restriction on r. You wrote:

"Prove that there exists an odd integer M such that for all real numbers r larger than
M ..."

It's asserting that we can always find some odd integer M which makes any real number bigger than M satisfy the relation 1/2r < 1/100.

My only concern is that this looks a tad on the trivial side. Is the exercise just in writing and interpreting the symbols?

 

[Aziza]

$$\forall r \in \mathbb{R} > M \; ; \; \exists M = 2n-1 \; \wedge \; n \in \mathbb{Z} \; : \; \frac{1}{2r} < \frac{1}{100}$$

but that's not the only restriction on r. You wrote:

"Prove that there exists an odd integer M such that for all real numbers r larger than
M ..."

It's asserting that we can always find some odd integer M which makes any real number bigger than M satisfy the relation 1/2r < 1/100.

My only concern is that this looks a tad on the trivial side. Is the exercise just in writing and interpreting the symbols?

Ohh true true...I didn't know you could write something like $$\forall r \in \mathbb{R} > M$$...is this a common notation? I mean, would a professor accept it? I am preparing for my transition to advanced math course this fall so this is a concern for me...my book always only uses one restriction in front of the "for all" or "there exists" symbol, so I thought that any additional restriction could be connected with an "and" to the rest of the statement...so to negate $\forall r \in \mathbb{R} > M$, it would just be:
$\exists r \in \mathbb{R} > M$ , right?

So the following statement
"For every positive real number x, there is a positive real number y less
than x with the property that for all positive real numbers z, yz ≥ z."
can be expressed as:

$(\forall x\in\mathbb{R^+})(\exists y\in\mathbb{R^+}<x)(\forall z\in\mathbb{R^+})(yz≥z)$
right?
and yea these proofs themselves are meant to be trivial, it's just about using basic proof techniques and knowing how to express statements using logical symbols

 

[Simon Bridge]

OK since it is the point of this section, you'd better stick to being pedantic for now. I've never been penalized for that sort of shorthand and, if you skim a bunch of journals you'll find that convention gets abandoned in favor or readability.

Sadly, what gets emphasized seems to depend on the school.

What I wanted to show you was that you can make more sense out of things by working backwards.