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

1. Jul 28, 2012

### 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 [Broken] 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 [Broken] 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...

Last edited by a moderator: May 6, 2017
2. Jul 28, 2012

### Simon Bridge

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

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?

Last edited by a moderator: May 6, 2017
3. Jul 28, 2012

### Aziza

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

4. Jul 29, 2012

### 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.