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Operations across an inequality

  • #1
rock.freak667
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Homework Statement



If a<b...can I then say that [itex]a^2<b^2[/itex] or [itex]a^3<b^3[/itex] or even [itex]\sqrt{a}<\sqrt{b}[/itex]?


Why I ask is that usually in some of the induction questions I do with inequalities, they give an expression like [itex]x_{n+1}=x_n^2 -7 [/itex] or something like that.
 

Answers and Replies

  • #2
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Consider cases where a, b are positive or negative.
 
  • #3
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also, another key thing to consider is greater than or less than 1.
 
  • #4
rock.freak667
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Consider cases where a, b are positive or negative.
Choose a=-1 and b=2, -1<2 true

[itex](-1)^2=1, (2)^2=4....1<4 [/itex] true.

a=2,b=3; 2<3 true

[itex] (2)^2=4,(3)^2=9....4<9[/itex] true

So I can do it across the inequality?
 
  • #5
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but say: -3<2 however (-3)^2=9 is not smaller than 2^2=4.
 
  • #6
rock.freak667
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but say: -3<2 however (-3)^2=9 is not smaller than 2^2=4.
ah right....sooo...then if am given something like

[tex]u_{n+1}=-1+\sqrt{u_n +7}[/itex]

and to prove [itex]u_n<2[/itex]

I can't add 7 to both sides and then take the square root as my inequality might not be the same after taking the square root? If so...then how would I manipulate the [itex]u_n[/itex] to get [itex]u_{n+1}[/itex]
 
  • #7
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Can you post the original problem, because i don't think i am getting what you are saying?

Are you asked to prove that u_n<2, or you are provided with this fact? Because if the former is the case then you cannot start from this, since you are actually asked to prove this. IN other words, what is the problem asking you to do?
 
  • #8
rock.freak667
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Can you post the original problem, because i don't think i am getting what you are saying?

Are you asked to prove that u_n<2, or you are provided with this fact? Because if the former is the case then you cannot start from this, since you are actually asked to prove this. IN other words, what is the problem asking you to do?
Prove by mathematical induction that [itex]u_n<2[/itex] given that [itex]u_{n+1}=-1+\sqrt{u_n +7}[/itex]
 
  • #9
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Prove by mathematical induction that [itex]u_n<2[/itex] given that [itex]u_{n+1}=-1+\sqrt{u_n +7}[/itex]
And you have no prior information about [tex] u_1,... or .... u_0....??[/tex] It looks kind of hard, without having at leas some other information about this sequence besides that recurrence relation.
 
  • #10
rock.freak667
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Sorry...[itex]u_1 =1[/itex]
 
  • #11
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Ok then,

[tex]u_{n+1}=-1+\sqrt{u_n +7}[/tex]


let n=1, so

[itex]u_{2}=-1+\sqrt{u_1 +7}=-1+\sqrt{8}=-1+2\sqrt{2}<2[/itex]

So for n=1, the inequality is valid.

Now let's suppose that the inequality is valid for n-1, that is

[tex] u_{n-1}<2--------(IH)[/tex] we need to show now that it is also true of n. That is we need to show that

[tex]u_n<2...???[/tex]

[tex]u_n=-1+\sqrt{u_{n-1}+7}[/tex]

now from IH we have [tex]u_{n-1}+7<2+7=9=>\sqrt{u_{n-1}+7}<\sqrt{9}=3----------(*****)[/tex] so


[tex]u_n=-1+\sqrt{u_{n-1}+7}<-1+\sqrt{9}=-1+3=2[/tex]

What we actually wanted to prove.!!!

P.S. Step (*****) is valid, because the sequence is increasing, so since u1=1, it means that [tex] u_n>0,\forall n[/tex]
 
Last edited:
  • #12
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So since u_n is a monotonic increasing sequence and bounded, its limit exists, so it also is a convergent sequence. If you are interested in finding the limit, it is very straightforward, just let:

[tex]\lim_{n\to\infty}u_n=L=\lim_{n\to\infty}u_{n+1}[/tex] and use the reccurrence relation, you will end up with two solutions, and you'll know which one to choose when you get there.

I hope i was of any help!!!!
 
  • #13
rock.freak667
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now from IH we have [tex]u_{n-1}+7<2+7=9=>\sqrt{u_{n-1}+7}<\sqrt{9}=3----------(*****)[/tex]

P.S. Step (*****) is valid, because the sequence is increasing, so since u1=1, it means that [tex] u_n>0,\forall n[/tex]
ahh..thanks...that is what I wanted to know. If and why it was valid to take the square root on both sides.
 
  • #14
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ahh..thanks...that is what I wanted to know. If and why it was valid to take the square root on both sides.
Just to be more precise, it should have read here:

[tex]u_n\geq 1,\forall n[/tex]
 
  • #15
tiny-tim
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If a<b...can I then say that [itex]a^2<b^2[/itex] or [itex]a^3<b^3[/itex] or even [itex]\sqrt{a}<\sqrt{b}[/itex]?
Hi rock.freak667! :smile:

Rewrite [itex]a^2<b^2[/itex] as [itex]a^2\,-\,b^2\,<\,0\,.[/itex]

Then (a - b)(a + b) < 0, so either a < b and a + b > 0, or a > b and a + b < 0.
Prove by mathematical induction that [itex]u_n<2[/itex] given that [itex]u_{n+1}=-1+\sqrt{u_n +7}[/itex]
"Prove by induction" means that you can assume that un < 2.

You'd better also assume that un > -1.

And √ in this case means the positive root.

Then √(un + 7) is between √6 and √9, so … ? :smile:
 
  • #16
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You'd better also assume that un > -1.
Well he need not assume this at all, since [tex] u_1=1[/tex] and u_n is an increasing sequence, it means like i said in one of my other posts, that

[tex] u_n\geq 1,\forall n \in N[/tex]

So all other opertations follow because of this.!!!!!
 

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