Determining if a number is within an interval

[rxh140630]

Homework Statement

I can't see itex formatting, and if I made a mistake in formatting or not, for the homework statement section, so as to save time and any confusion I will write the question in the solution attempt.

Relevant Equations

none that I know

0<k<1
x<y
$x,y \in {[a,b]}$
$a,b \in {\mathbb R}$

Question: is $yk + (1-k)x \in {[a,b]}$

My response:

$yk + (1-k)y = y$

Since $x<y$, $yk + (1-k)x < y$

$xk + (1-k)x = x$

Since $y>x$, $yk + (1-k)x > x$

Therefore $x < yk + (1-k)x < y$, so yk + (1-k)x is in the interval [a,b]

Is this considered proven? Did I miss anything?

 

[member 587159]

$yk + (1-k)y =y$ makes no sense. This is not true for all values of $k$. You must also require $a < b$. Your goal is to show that $yk + (1-k)x \in [a,b]$ and you showed that $yk + (1-k)x \in [x,y] \subseteq [a,b]$ so I guess your attempt does contain the right idea, but the exposition can be better imo.

 

[rxh140630]

$yk + (1-k)y =y$ makes no sense. This is not true for all values of $k$. You must also require $a < b$. Your goal is to show that $yk + (1-k)x \in [a,b]$ and you showed that $yk + (1-k)x \in [x,y] \subseteq [a,b]$ so I guess your attempt does contain the right idea, but the exposition can be better imo.

Sorry, yes also a<b. I don't understand why $yk + (1-k)y = y$ doesn't make sense? For all values of k s.t 0<k<1, yk + (1-k)y = yk + y -yk = y.

 

[member 587159]

Sorry, yes also a<b. I don't understand why $yk + (1-k)y = y$ doesn't make sense? For all values of k s.t 0<k<1, yk + (1-k)y = yk + y -yk = y.

I misread and misquoted sorry. It surely is correct. But that sentence is not necessary for your proof?

 

[rxh140630]

Perhaps not, just wanted to make sure that it's correct, regardless of elegance.

 

[FactChecker]

Your proof is valid. I would just encourage you to make it a little more step-by-step, like this:
Since x < y and k > 0, $xk + (1-k)y < yk + (1-k)y = y < b$.