Operations across an inequality

In summary, we discussed the use of inequalities in mathematical induction, and how to manipulate them using specific cases and the monotonicity of a sequence. We also discussed how to use the assumption of the induction hypothesis to prove the desired inequality in the given recurrence relation.
  • #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.
 
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  • #2
Consider cases where a, b are positive or negative.
 
  • #3
also, another key thing to consider is greater than or less than 1.
 
  • #4
mutton said:
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
but say: -3<2 however (-3)^2=9 is not smaller than 2^2=4.
 
  • #6
sutupidmath said:
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
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
sutupidmath said:
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
rock.freak667 said:
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
Sorry...[itex]u_1 =1[/itex]
 
  • #11
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]
 
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  • #12
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
sutupidmath said:
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
rock.freak667 said:
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
rock.freak667 said:
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.
rock.freak667 said:
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
tiny-tim said:
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.!
 

What is an inequality?

An inequality is a mathematical statement that compares two values or quantities and shows the relationship between them using symbols such as <, >, ≤, ≥. It indicates that one value is less than, greater than, less than or equal to, or greater than or equal to the other value.

What are the operations that can be performed across an inequality?

The operations that can be performed across an inequality are addition, subtraction, multiplication, and division. These operations affect the inequality sign and must be done to both sides of the inequality to maintain its balance.

When do we reverse the inequality sign when performing operations?

The inequality sign is reversed when multiplying or dividing both sides of the inequality by a negative number. This is because multiplying or dividing by a negative number changes the direction of the inequality. For example, if we have 5 < 10 and we multiply both sides by -1, we get -5 > -10.

Do we always perform the same operation on both sides of an inequality?

Yes, we always perform the same operation on both sides of an inequality. This is to maintain the balance of the inequality. If we add a value to one side, we must also add the same value to the other side. If we subtract a value from one side, we must also subtract the same value from the other side.

Can we simplify an inequality using operations?

Yes, we can simplify an inequality using operations. This is done to make the inequality easier to solve or to find a specific solution. We can use the properties of inequality to combine like terms, move terms to the same side, or eliminate terms to simplify the inequality.

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