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$ x,y,a\; \in F $ which is an ordered field and this is given:

$$ 0 \lt a \lt 1 $$ $$ x \lt y $$

I need to prove:

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

This is a relatively simple proof, yet I'm having quite the difficulty with it. What's the strategy for attacking proofs? Brute forcing it (i.e. trying many different steps without too much thought isn't working) isn't working well for me.

I've tried starting with $ 0 \lt a \lt 1 $ and doing ordered field operations (adding to each side, multiplying by each side etc) but I'm unable to get anywhere close to an actual proof.

So I've done this:

$$ 0\lt a \lt 1$$

Multiply sides by (x-y):

$$ 0*(x-y)\lt a(x-y) \lt (x-y)*1 $$ $$ 0\lt a(x-y) \lt (x-y)$$

Adding y:

$$ y+ 0\lt a(x-y)+y \lt x-y+y $$

$$ y\lt a(x-y)+y \lt x $$

Here is where I appear to have difficulties continuing.

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    Hint: $ax + (1-a)y = y + a(x - y)$, and $ax + (1-a)y = (1-a')x + a'y = x + a'(y -x)$ where $a' = 1-a$. – Eric Thoma Feb 29 '16 at 19:09
  • Thanks for the reply @EricThoma. I won't be able to use stackexchange during exams so I need to get some sort of intuition as what steps to take in order to solve proofs. –  Feb 29 '16 at 19:11
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    There is a geometric picture for this proof. Let $x,y \in \mathbb{R}$ and draw the points $x$, $y$, and $ax + (1-a)y$ on a number line. Are you asking about how to approach proofs in general? Or proofs in a specific area of math? – Eric Thoma Feb 29 '16 at 19:14
  • @EricThoma Proofs in general. Is there some sort of general algorithm ? –  Feb 29 '16 at 19:15
  • Unfortunately (or maybe fortunately depending on who you ask), there is nothing quite so easy. You may find the answers to these questions helpful: #1, #2. – Eric Thoma Feb 29 '16 at 19:19
  • @EricThoma thank you for your time! I've updated the original question with some work, but I'm stuck near the end. –  Feb 29 '16 at 19:27
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    When you multiply by $x-y$, you are multiplying by a negative element, so you have to reverse inequality signs. The general relation $s + r < s$ when $r < 0$ holds, for any $s,r \in F$, so see if you can apply this to the expression $y + a(x-y)$ to see that it is less than $y$. – Eric Thoma Feb 29 '16 at 19:31
  • @EricThoma Of course!! Thank you! –  Feb 29 '16 at 19:33

1 Answers1

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There's no step-by-step algorithm for proofs in general, as far as I'm aware; even if it's limited to "proofs I am likely to be asked for on the next exam" I suspect it would be too difficult to devise (and learn) a method that is guaranteed to work in every case.

It may be possible to gain a little more facility in what you try when approaching a problem, however. For example, when I saw this problem, just about the first thought that came to mind was, "There's a single $y$ on the far right-hand side; what do I get if I cancel it? Can I prove the resulting inequality?"

Indeed, if we subtract $y$ from each side of $ax+(1−a)y<y$, we get $$ ax - ay < 0, $$ which seems easy enough to prove, since $x < y$ and $a > 0$.

Then I started to try the same thing for the $x$ on the far left, but then I wondered which makes more sense, to cancel the $x$ on the left of $x < ax + (1-a)y$, or to cancel the $ax$ on the right? It turns out that if we cancel $x$ we get $0 < (a - 1)x + (1-a)y$, which is provable, but if we cancel $ax$ we get directly to $(1-a)x < (1 - a)y$, which seems simpler.

It then occurred to me that maybe it's nicer to cancel $(1-a)y$ rather than $y$ on the right, and indeed that gets us $ax < ay$.

But whether you chose the "nice" way or the "less nice" way, the task would then be to prove the new fact and use it to prove the original fact. The pattern seems to be, for each of the inequalities in the original statement, you start with $x<y$, then multiply by either $a$ or $1-a$ (either one is positive, so it preserves the $<$), possibly collect terms on one side of the $>$, then add the necessary quantity, $ax$, $(1-a)y$, $x$, or $y$ (depending on which of the two inequalities you're working on and the proof path you chose) in order to achieve the desired inequality.

David K
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  • Thank you for sharing your perspective on the problem David, it definitely helped! –  Feb 29 '16 at 20:19