Proof by Contradiction: Showing x ≤ 1 for x∈ℝ+ and t∈T

[yy205001]

Homework Statement

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$\leq$x^t 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.

Homework Equations

The Attempt at a Solution

First, I assume the theorem is false, so negation of If x$\leq$x^t then x$\leq$1 is true.

The negation of the theorem is: x$\leq$x^t $\wedge$ x>1
x$\leq$x^t $\wedge$ x>1 Premis
x$\leq$x^t Inference rule for conjunction
log(x) ≤ log(x^t) 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$\leq$x^t then x$\leq$1 is false
Thus, If x$\leq$x^t then x$\leq$1

 

[Joffan]

What is the sign of log(x)?

 

[yy205001]

What is the sign of log(x)?

It is positive.

 

[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.

 

[yy205001]

Thank you aleph-aleph for helping me on this!