1. Apr 13, 2013

### yy205001

1. The problem statement, all variables and given/known data
Please check that the proof is correct or not.
Let ℝ+ = {x$\inℝ$: x>0} and T = {x$\inℝ$: 0<x<1}.

Let x∈ℝ+ and t∈T

Prove: If x$\leq$xt then x$\leq$1.

* You may assume any common properties of log(x) as well as : if 0<a$\leq b$ then log(a) ≤ log(b)

Any help is appreciated.

2. Relevant equations

3. The attempt at a solution
First, I assume the theorem is false, so negation of If x$\leq$xt then x$\leq$1 is true.

The negation of the theorem is: x$\leq$xt $\wedge$ x>1
x$\leq$xt $\wedge$ x>1 Premis
x$\leq$xt Inference rule for conjunction
log(x) ≤ log(xt) log both side
log(x) ≤ t*log(x) properties of log
1 ≤ t
which is a contradiction with the domain of t since 0<t<1

Therefore, the negation of If x$\leq$xt then x$\leq$1 is false
Thus, If x$\leq$xt then x$\leq$1

Last edited: Apr 13, 2013
2. Apr 13, 2013

### Joffan

What is the sign of log(x)?

3. Apr 13, 2013

### yy205001

It is positive.

4. Apr 14, 2013

### aleph-aleph

The negation of the statement should be x≤xt => x>1 instead of x≤xt ∧ x>1.
Other than that, I believe your proof is valid.

5. Apr 15, 2013

### yy205001

Thank you aleph-aleph for helping me on this!

Draft saved Draft deleted

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