My Math Forum proof involving real number!

 Real Analysis Real Analysis Math Forum

 March 10th, 2013, 07:51 PM #1 Member   Joined: Sep 2012 Posts: 69 Thanks: 0 proof involving real number! Question Let x be a positive real number. Prove that if $x-\frac{2}{x} > 1$, then $x>2$. (a) a direct proof Assume $x-\frac{2}{x} > 1$ is true. Then, $x^2 -2 > x$ $x^2 -x > 2$ $x(x-1) > 2$ Since $x \in R^+$ and $x(x-1)>2$, $x-1$ is positive real number. If x is less than or equal to 2, then $x(x-1) \leq 2$. Therefore, $x>2$. Q.E.D Any comment?? And can anyone prove this by contrapositive?? Thanks.
 March 10th, 2013, 09:30 PM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: proof involving real number! Correct. Maybe you should have stated the reason for x(x-2)<=2. Anyway. x^2-x = (x-1/2)^2 - 1/4 you can wtite (x-1/2)^2 > 9/4 i.e. x>2 As we can see the (in)equality x-2/x ? 1 is logically the same than x ? 2.
 March 12th, 2013, 02:41 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: proof involving real number! Contrapositive: If $0< x\le 2$ then $\frac{2}{x}\ge 1$ so $x- \frac{1}{x}=> 2- 1= 2$.

 Tags involving, number, proof, real

,

,

# proof involving numbers

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post rnck Advanced Statistics 6 July 1st, 2013 06:21 AM Michaeld71 Algebra 3 November 25th, 2012 03:17 PM TsAmE Complex Analysis 1 October 18th, 2010 04:38 PM RastaMasta Algebra 3 October 20th, 2009 12:25 PM eChung00 Applied Math 3 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top