Need help with this induction proof

  • Thread starter garyljc
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  • #26
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As noted before, have you already proved "if x< y and 0< a< b, then ax< by"?
is anyone here ?
help me
i've post the proof

halls , i was wondering if i should just print this "if x< y and 0< a< b, then ax< by" since it's rather trivial
it's a one line proof
 
  • #27
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Have you already proved that: it a< b and x< y, then ax< by?
this is very trivial
do we have to proof it ?
if so , how do we ?
 
  • #28
CompuChip
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Well, how trivial it is depends on how precise you want to get. I could also say that what you want to prove is trivial and you don't need to prove it, but apparently you think otherwise :smile:
Currently, your proof is globally of the form:
"Suppose <induction hypothesis>, then <what we want to prove> is true so we have proven <what we want to prove>".

So you can assume this axiom (as said before)
If x<y, then xz<yz, for any z>0​
Note that on both sides you are multiplying by the same number. So the axiom does not say that:
If 0<x<y, then x^2<y^2​
because z would have to be equal to x and y at the same time. So first try to find the missing step(s) here.
 
  • #29
HallsofIvy
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is anyone here ?
help me
i've post the proof

halls , i was wondering if i should just print this "if x< y and 0< a< b, then ax< by" since it's rather trivial
it's a one line proof
this is very trivial
do we have to proof it ?
if so , how do we ?
You've said twice now that it is "trivial" and once that it is "a one line proof". Why are you now asking HOW to prove it?

Actually, the statement, as given here, is NOT true. For example, if a= 1, b= 2, x= -3, y= -2 then x< y, 0<a< b but ax= -3, by= -4 so ax is NOT less than by. You need at least y> 0 also. In that case, since x< y and a> 0, ax< ay. Since a< b and y> 0, ay< by. From ax< ay< by, ax< by.
 

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