## Mathematical Induction

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

Consider the sequence of real numbers x1, x2, x3,.... defi ned by the relations x1 = 1

and xn+1 =$$\sqrt{1 + 2xn}$$

1. Use mathematical induction to show that xn+1 > xn
for all n $$\geq$$ 1.

3. The attempt at a solution

I'm a bit thrown off by this question because it seems very obvious that it would be greater. if x1=1 is it safe to assume that xn=n? and if so I feel like this proof is stupid cause it shows right in front of you that its greater. any help is great thanks.
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Mentor
 Quote by AerospaceEng 1. The problem statement, all variables and given/known data Consider the sequence of real numbers x1, x2, x3,.... defi ned by the relations x1 = 1 and xn+1 =$$\sqrt{1 + 2x_n}$$ 1. Use mathematical induction to show that xn+1 > xn for all n $$\geq$$ 1. 3. The attempt at a solution I'm a bit thrown off by this question because it seems very obvious that it would be greater. if x1=1 is it safe to assume that xn=n? and if so I feel like this proof is stupid cause it shows right in front of you that its greater. any help is great thanks.
The obviousness (or not) is irrelevant to what you need to do, which is to prove this statement by induction.

You are given x1 = 1. What is the value for x2? If x2 > x1, then use that for your base case.

Next, assume that xk > xk - 1 (the induction hypothesis), and use that assumption to show that xk + 1 > xk. If you can do this, you will have proved that the statement is true for all n >= 1.
 The proof is not stupid! Lol engineers! Perform the substraction $$x_{n+1}- x_n$$ then show the substraction is > 0. Hint: rationalise and use the fact that a^2 >0 for an real number a.

## Mathematical Induction

I'm I allowed to say that xn+1 is the same thing as xn + x1 cause i feel I can't and if I can't I feel so limited.

Recognitions:
Gold Member
 Quote by AerospaceEng I'm I allowed to say that xn+1 is the same thing as xn + x1 cause i feel I can't and if I can't I feel so limited.
But that's not true.

xn+1 is clearly shown to be $$\sqrt{1+2x_n}$$.

So...

$$x_{n+1}-x_n = \sqrt{1+2x_n} - x_n$$

So you need to prove that the above equation is always greater than 0, given x1=1.
 okay so I think i see where this is going. what i've done so far is i've taken the RHS and multiplied it by itself just with a positive between the terms, to give me 1+2xk-xk2 But i feel like i've done something wrong because i haven't used x1=1 as of yet..

Recognitions:
Gold Member
 Quote by AerospaceEng okay so I think i see where this is going. what i've done so far is i've taken the LHS and multiplied it by itself just with a positive between the terms, to give me 1+2xk+xk2 and i can factor this to give me (x+1)2 Is that good enough to show that it's always greater than 0. and then ill just plug this mini proof into my main proof. But i feel like i've done something wrong because i haven't used x1=1 as of yet..
Considering that x_k itself is always greater than 0, that should be enough. But I'll wait for a second opinion on that.
 now if i work with the left hand side I get xk+12-xk2 and i can cancel the -xk2 on both sides to give me: xk+12=2k+1 and i feel im stuck again and where does the x1=1 come into play?
 Recognitions: Homework Help the x_1 comes into play when you show that x_2>x_1 which is the initial step when you should have done first.

Mentor
 Quote by AerospaceEng now if i work with the left hand side I get xk+12-xk2 and i can cancel the -xk2 on both sides to give me: xk+12=2k+1 and i feel im stuck again and where does the x1=1 come into play?
See post #2.

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