Proof Exercise II

  • #1
reenmachine
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Homework Statement



Suppose ##x## and ##y## are positive real numbers.If ##x < y## , then ##x^2 < y^2##.

Homework Equations



Assume ##x<y## with ##x,y \in ℝ^+##.This implies that ##\exists z \in ℝ^+## such that ##x+z=y##.We have ##y^2 = (x+z)^2 = x^2 + 2xz + z^2## , and this proves that if ##x < y## with ##x,y \in ℝ^+## , then ##x^2 < y^2##.

any thoughts on that one? Thank you!
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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hi reenmachine! :wink:

yes, that looks fine :smile:

(alternatively, you could factor y2 - x2)
 
  • #3
hilbert2
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You only need to know that if ##x>0##, ##y>0## and ##x<y## then ##x+y>0## and ##x-y<0##.

Also, if ##a>0## and ##b<0## then ##ab<0##.
 
  • #4
pasmith
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You need to explain why [itex]2xz + z^2[/itex] is strictly positive. It is obvious, but then so is the result you're trying to prove.

An alternative proof is that if [itex]0 < x < y[/itex] then, since multiplying by a positive number preserves inequalities, we have [itex]x^2 < xy[/itex] on multiplying by [itex]x[/itex] and [itex]xy < y^2[/itex] on multiplying by [itex]y[/itex]. Putting these together we have [itex]x^2 < xy < y^2[/itex].
 
  • #5
reenmachine
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hi reenmachine! :wink:

yes, that looks fine :smile:

(alternatively, you could factor y2 - x2)
Hi! :)

thank you for taking the time to respond!
 
  • #6
reenmachine
Gold Member
513
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You need to explain why [itex]2xz + z^2[/itex] is strictly positive. It is obvious, but then so is the result you're trying to prove.

An alternative proof is that if [itex]0 < x < y[/itex] then, since multiplying by a positive number preserves inequalities, we have [itex]x^2 < xy[/itex] on multiplying by [itex]x[/itex] and [itex]xy < y^2[/itex] on multiplying by [itex]y[/itex]. Putting these together we have [itex]x^2 < xy < y^2[/itex].
I thought the fact that [itex]2xz + z^2[/itex] is positive was obvious from the fact x and z are in ##R^+##...

I like your way of doing it , pretty good road to the proof.

thank you!
 

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