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So I have this one homework assignment where I have to prove the following clause "If $x+y \ge 1$ then $x \ge \frac 12$ or $y \ge \frac 12$". I thought that if I assign $x=y$ and put it like "$2x \ge 1$" and solve the x, does it actually prove that either of them(or both) need to be greater than 1/2?

The clause "If $x + y ≥ 1$ then $x ≥ 1/2$ or $y ≥ 1/2$"

Edit x,y are real numbers

Sorry if this is too simple to post here or something but I've always had problems with the proving assignments which I really need to start learning as I started in university like two weeks ago. I am not asking for complete solution, tips are ok.

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    $\begingroup$ If $x,y<\frac12, x+y<1$ $\endgroup$ – lab bhattacharjee Sep 11 '13 at 18:15
  • $\begingroup$ I agree with @labbhattacharjee. Argue by contraposition. $\endgroup$ – Giuseppe Negro Sep 11 '13 at 18:20
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    $\begingroup$ Without loss of generality, suppose that $x \geqslant y$. Then $2x \geqslant x + y \dotsc$ $\endgroup$ – Daniel Fischer Sep 11 '13 at 18:20
  • $\begingroup$ @GiuseppeNegro, a more logical statement is at least one of $x,y$ is $\ge\frac12$ if $x+y\ge1$ $\endgroup$ – lab bhattacharjee Sep 11 '13 at 18:30
  • $\begingroup$ @labbhattacharjee Isn't the negation of the If "x + y ≥ 1" then x ≥ 1/2 or y ≥ 1/2" clause "There are real numbers x and y, x ≥ 1/2 or y ≥ 1/2 if x+y < 1 ? Or did I get it completely wrong. $\endgroup$ – Samuli Lehtonen Sep 11 '13 at 18:44
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HINT: Suppose that it’s not true, i.e., that $x<\frac12$ and $y<\frac12$; what can you then say about $x+y$?

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  • $\begingroup$ Then x + y ≥ 1 can't be true so the negation is wrong which makes the original statement true, but I don't really get how I should write it on the paper the part where I show that x + y ≥ 1 is false? Do I assign the x=y and show it that way? $\endgroup$ – Samuli Lehtonen Sep 11 '13 at 19:08
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    $\begingroup$ @Samuli: You could say something like this: If $x<\frac12$ and $y<\frac12$, then $x+y<\frac12+\frac12=1$, contradicting the hypothesis that $x+y\ge 1$. Therefore it’s not the case that both $x$ and $y$ are less than $\frac12$, so either $x\ge\frac12$ or $y\ge\frac12$ (or both). (You don’t actually need proof by contradiction here: this can also be phrased as a proof of the contrapositive. The contrapositive of a statement p implies q is not-p implies not-q; the two are logically equivalent, so proving one proves the other.) $\endgroup$ – Brian M. Scott Sep 11 '13 at 19:15
  • $\begingroup$ Yeah. We haven't dealt with the proof of the contrapositive yet in class so I thought the teacher would assume that we'd use the contradiction method but I guess it's fine to use the contrapositive too. Is it correct too if I do it with the contradiction method like I said? $\endgroup$ – Samuli Lehtonen Sep 11 '13 at 19:20
  • $\begingroup$ @Samuli: Yes, it is. (I thought that contradiction might be more familiar; that’s why I wrote it up that way.) $\endgroup$ – Brian M. Scott Sep 11 '13 at 19:22
  • $\begingroup$ Ok, great! I understand now, thanks! $\endgroup$ – Samuli Lehtonen Sep 11 '13 at 19:45
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You have $\max(x,y)+\min(x,y) \ge 1$. Since $\max(x,y)\ge\min(x,y)$, this becomes $2 \max(x,y) \ge 1$, or in other words, $$ \max(x,y) \ge \frac{1}{2}$$

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