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
[itex] x,y \in {[a,b]} [/itex]
[itex]a,b \in {\mathbb R}[/itex]

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

My response:

[itex] yk + (1-k)y = y [/itex]

Since [itex] x<y[/itex], [itex] yk + (1-k)x < y [/itex]

[itex]xk + (1-k)x = x [/itex]

Since [itex] y>x [/itex], [itex]yk + (1-k)x > x[/itex]

Therefore [itex] x < yk + (1-k)x < y[/itex], 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 [itex] yk + (1-k)y = y [/itex] 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 [itex] yk + (1-k)y = y [/itex] 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##.