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If $0 \le a \le 1$, then show that $xa + (1-a)y$ will always lie between $x$ and $y$.

I am sorry if this may seem like elementary question. I have tried many examples and they all seem to work. Still I am not able to prove this.

Can any one prove(or point me in some direction) this both intuitively and formally.

Umberto P.
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Abhinav Garg
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    actually it is between $y$ and $x$ as you wrote it. Just write $z=xa+(1-a)y=y + a(x-y)$ and it will be clear why it is like that. – G Cab Jun 14 '16 at 16:54
  • By proving this, How can you prove that e will be on line pq. While trying many examples I just happened to observe that it generalizes to all and then I asked this question. I then realized that it is now a mathematical question and thus I posted here. – Abhinav Garg Jun 14 '16 at 17:19
  • If we consider $x,y$ as data, then the value $xa + (1-a)y$ for $a \in [0,1]$ corresponds to a weighted average of this data. Averages are always in between data. – MathematicsStudent1122 Jun 14 '16 at 17:30
  • @AbhinavGarg, my mistake. They're not the same. Sorry about that. – D.W. Jun 14 '16 at 18:28
  • @D.W. Its Ok. You were only performing your duty. Mistakes Happen. We are only humans. So now can you please remove the duplicate tag? It might hinder the question's visibility. – Abhinav Garg Jun 14 '16 at 19:23

4 Answers4

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Assuming that $y\le x$ $$\text{Let }0\le a \le 1.$$ $$\implies 0\le a(x-y) \le x-y$$ $$\implies y\le y+ a(x-y) \le y+x-y$$ $$\implies y\le ax+(1-a)y \le x$$

If $x\lt y$, the same argument applies by setting $b=1-a$, thus $$x\le by +(1-b)x\lt y$$

flawr
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John Joy
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Assuming $\;x<y\;$ :

$$\begin{align*}&ax+(1-a)y\le a\color{red}y+(1-a)y=y\\{}\\ {}\\&ax+(1-a)y\ge ax+(1-a)\color{blue}x=x\end{align*}$$

Thus, from the both lines above

$$x\le ax+(1-a)y\le y$$

DonAntonio
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HINT

Notice you have a continuous monotone function with $f(0) = y$ and $f(1)=x$...

Emily
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gt6989b
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  • This question is marked pre-calculus though. – I. J. Kennedy Jun 14 '16 at 16:56
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    @I.J.Kennedy what do you need calculus for? $f$ is linear in $a$ and hence is monotone, no derivatives necessary. – gt6989b Jun 14 '16 at 16:57
  • @gt6989b I think Kennedy may have meant that you need continuity and the intermediate value theorem, which is usually taught in calculus, not in pre-calculus. – DonAntonio Jun 14 '16 at 17:50
  • @Joanpemo with effort you can replace continuity and linearity... formally you do need the IVT for this argument, i agree... – gt6989b Jun 14 '16 at 18:04
  • It is particularly desirable when using (relative to the Question) advanced ideas to solve a problem, to make their presentation more of a coherent argument and less of an abrupt "hint". – hardmath Jun 14 '16 at 19:54
  • I know a little bit of elementary calculus ideas like continuity, what do derivatives and integrals signify. If you can then please solve it using calculus. If there are any advanced ideas I can very well learn them – Abhinav Garg Jun 14 '16 at 20:09
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It is not restrictive to assume $x\le y$. Consider $$ xa+(1-a)y=xa+x(1-a)-x(1-a)+(1-a)y= x+(1-a)(y-x) $$ and conclude $xa+(1-a)y\ge x$.

Can you do similarly for getting $xa+(1-a)y\le y$?

egreg
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