Can someone tell me why this inequality proof is wrong?

  • Thread starter nate9228
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  • #1
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Homework Statement


Let x and y be real numbers such that x<y. There exists z that is a real number such that x<z<y.


Homework Equations





The Attempt at a Solution


I wrote the following: We are given that x<y. We know from previous proof that (1/2)<1.
Consider y such that y=x+1
x+1<x and we know (1/2)<1. Therefore x+(1/2)<x+1
Also, x<x+(1/2). Therefore x<x+(1/2)<x+1 which shows the existence of a z such that x<z<y.

For this problem my professor marked me wrong at the part where I said let y=x+1. He just draws a red line and leaves it up to us to figure what we did wrong. I do not see how this argument is invalid though. Can someone explain?
 

Answers and Replies

  • #2
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Correction in line 3 under my solution. I meant to type x+1>x
 
  • #3
Dick
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Homework Statement


Let x and y be real numbers such that x<y. There exists z that is a real number such that x<z<y.


Homework Equations





The Attempt at a Solution


I wrote the following: We are given that x<y. We know from previous proof that (1/2)<1.
Consider y such that y=x+1
x+1<x and we know (1/2)<1. Therefore x+(1/2)<x+1
Also, x<x+(1/2). Therefore x<x+(1/2)<x+1 which shows the existence of a z such that x<z<y.

For this problem my professor marked me wrong at the part where I said let y=x+1. He just draws a red line and leaves it up to us to figure what we did wrong. I do not see how this argument is invalid though. Can someone explain?
The argument is valid. It just only works if y=x+1. That really a special case of what you want to show. Suppose y isn't equal to x+1?
 
  • #4
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Hi nate,
well your teacher is right, x and y are already "given"
why would you say let y=x+1 ?
y can be anything, it is just > x, but with an arbitrarily small or big difference (it could be x+1000000000, or x+0.000000001 or whatever, you just know y>x).
since you are talking about real numbers, you know that (x+y)/2 exists and is right between x and y

Cheers...
 
  • #5
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I had my suspicions that that was why Dick, but I wanted to get another opinion, so thank you. Same to Oli4. In reference to Oli's post would this work:
You have a given x,y such that x<y. Adding x to both sides you have
x+x<y+x which implies 2x<y+x. Now divide by 2 and you have x<(y+x)/2.
Similarly, taking x<y and adding y you gave x+y<2y and again dividing by 2 yields (x+y)/2<y. Therefore you have x<(x+y)/2<y. Other then the non-formal way in which I typed everything, this is what Oli was referring to if I am not mistaken?
 
  • #6
Dick
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I had my suspicions that that was why Dick, but I wanted to get another opinion, so thank you. Same to Oli4. In reference to Oli's post would this work:
You have a given x,y such that x<y. Adding x to both sides you have
x+x<y+x which implies 2x<y+x. Now divide by 2 and you have x<(y+x)/2.
Similarly, taking x<y and adding y you gave x+y<2y and again dividing by 2 yields (x+y)/2<y. Therefore you have x<(x+y)/2<y. Other then the non-formal way in which I typed everything, this is what Oli was referring to if I am not mistaken?
Now that works for all y, not just y=x+1. Good job!
 
  • #7
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Yup, you got it, congrats :)
 
  • #8
42
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Awesome! Thanks :)
 

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