# Operations across an inequality

1. May 2, 2008

### rock.freak667

1. The problem statement, all variables and given/known data

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

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

2. May 2, 2008

### mutton

Consider cases where a, b are positive or negative.

3. May 2, 2008

### lzkelley

also, another key thing to consider is greater than or less than 1.

4. May 2, 2008

### rock.freak667

Choose a=-1 and b=2, -1<2 true

$(-1)^2=1, (2)^2=4....1<4$ true.

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

$(2)^2=4,(3)^2=9....4<9$ true

So I can do it across the inequality?

5. May 2, 2008

### sutupidmath

but say: -3<2 however (-3)^2=9 is not smaller than 2^2=4.

6. May 2, 2008

### rock.freak667

ah right....sooo...then if am given something like

$$u_{n+1}=-1+\sqrt{u_n +7}[/itex] and to prove $u_n<2$ I can't add 7 to both sides and then take the square root as my inequality might not be the same after taking the square root? If so...then how would I manipulate the $u_n$ to get $u_{n+1}$ 7. May 2, 2008 ### sutupidmath Can you post the original problem, because i don't think i am getting what you are saying? Are you asked to prove that u_n<2, or you are provided with this fact? Because if the former is the case then you cannot start from this, since you are actually asked to prove this. IN other words, what is the problem asking you to do? 8. May 2, 2008 ### rock.freak667 Prove by mathematical induction that $u_n<2$ given that $u_{n+1}=-1+\sqrt{u_n +7}$ 9. May 2, 2008 ### sutupidmath And you have no prior information about [tex] u_1,... or .... u_0....??$$ It looks kind of hard, without having at leas some other information about this sequence besides that recurrence relation.

10. May 2, 2008

### rock.freak667

Sorry...$u_1 =1$

11. May 2, 2008

### sutupidmath

Ok then,

$$u_{n+1}=-1+\sqrt{u_n +7}$$

let n=1, so

$u_{2}=-1+\sqrt{u_1 +7}=-1+\sqrt{8}=-1+2\sqrt{2}<2$

So for n=1, the inequality is valid.

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

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

$$u_n<2...???$$

$$u_n=-1+\sqrt{u_{n-1}+7}$$

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

$$u_n=-1+\sqrt{u_{n-1}+7}<-1+\sqrt{9}=-1+3=2$$

What we actually wanted to prove.!!!

P.S. Step (*****) is valid, because the sequence is increasing, so since u1=1, it means that $$u_n>0,\forall n$$

Last edited: May 2, 2008
12. May 2, 2008

### sutupidmath

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:

$$\lim_{n\to\infty}u_n=L=\lim_{n\to\infty}u_{n+1}$$ 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. May 2, 2008

### rock.freak667

ahh..thanks...that is what I wanted to know. If and why it was valid to take the square root on both sides.

14. May 2, 2008

### sutupidmath

Just to be more precise, it should have read here:

$$u_n\geq 1,\forall n$$

15. May 3, 2008

### tiny-tim

Hi rock.freak667!

Rewrite $a^2<b^2$ as $a^2\,-\,b^2\,<\,0\,.$

Then (a - b)(a + b) < 0, so either a < b and a + b > 0, or a > b and a + b < 0.
"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 … ?

16. May 3, 2008

### sutupidmath

Well he need not assume this at all, since $$u_1=1$$ and u_n is an increasing sequence, it means like i said in one of my other posts, that

$$u_n\geq 1,\forall n \in N$$

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