anonymous
  • anonymous
Let x be a real number. Prove if x>1 then x^2>x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Induction might work here.
OakTree
  • OakTree
Do you know what induction is, @zonazoo?
anonymous
  • anonymous
yes, but I don't see how you would use that here.

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OakTree
  • OakTree
Okay, so let's try a different approach. Let's look at the behavior of x^2 for all x>1. Do you see anything useful you could use there?
anonymous
  • anonymous
ummm, well, im not sure what you are looking for. if you square any number that is greater than 1, the answer will always be greater than the number you squared... the y value of the graph will always be greater on the quadratic than the linear function past 1. I would of thought you need to use contradiction, like suppose x
OakTree
  • OakTree
Contradiction would be a great way to prove this. Again, induction would be best, but contradiction would only be a line or two longer.
anonymous
  • anonymous
im not sure how you would do either.
OakTree
  • OakTree
Okay. So we have \[x^2 - x < 0\]So \[x(x-1) < 0\]But this means that exactly one of x or x-1 must be negative, since two negatives is positive and two positives is positive. Following so far?
anonymous
  • anonymous
yes... and this method would be by contradiction correct.
RadEn
  • RadEn
x > 1 multiply by x on both sides, we get x^2 > x :)
OakTree
  • OakTree
Right. Because we're assuming that something is wrong at the beginning. Anyway, so can you think of what the next step might be?
anonymous
  • anonymous
idk, just stating that either x<0 or x<1, which would be a contradiction since x>1???
OakTree
  • OakTree
Kind of. Since exactly one of x, x-1 must be negative, we know that x-1 is -1 and x is 0. But that means that x is not greater than 1. Contradiction. QED. Get it?
anonymous
  • anonymous
yes I do. What is QED?
OakTree
  • OakTree
Quod erat demonstrandum. It's Latin for "what was required to be proved" - it's basically a formal way of saying that the proof is finished.
anonymous
  • anonymous
oh okay. Thank you for all the help.
OakTree
  • OakTree
No problem.
anonymous
  • anonymous
I suppose induction may not be as easy as I had thought... the base case, namely, is giving me trouble. But if you restrict the domain here to be the natural numbers, it's easier to work out. Base case: \(x=2\). You have \(2^2=4>2\), so the inequality holds. Assume it holds for \(x=k\), so that \(k^2>k\). Next, show it holds for \(x=k+1\), i.e. \((k+1)^2>k+1\). The trick here is to write this inequality in terms of that given by the previous assumption: \[(k+1)^2=\color{red}{k^2}+\color{blue}{2k+1}>\color{red}{k}+\color{blue}{1}\] You know the red part on the left is greater than the red on the right, so now it's a matter of showing \(2k+1>1\), which is true for all \(k>0\). So the inequality holds for \(x=k+1\). But again, I'm not sure how one would adjust this for the real number system.

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