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Square root of a square plus 1

  • Thread starter icantadd
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  • #1
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Homework Statement


prove that if n is a natural number greater than 0, then sqrt(n^2 + 1) is not a natural number.


Homework Equations





The Attempt at a Solution


I can't tell if I am right, which probably means that I am not.

Assume sqrt(n^2+1) is a natural number. Then there is a natural number j such that j = sqrt(n^2+1). Thus
j^2 = n^2 + 1. Then, I am stuck.
It seems too easy to prove, is there some little trick I am missing.
 

Answers and Replies

  • #2
D H
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What is (j+1)2?
 
  • #3
HallsofIvy
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If [itex]\sqrt{n^2+1}= m[/itex] then [itex]n^2+ 1= m^2[/itex]. Now, what can you say about [itex]n^2- m^2[/itex]?
 
  • #4
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If [itex]\sqrt{n^2+1}= m[/itex] then [itex]n^2+ 1= m^2[/itex]. Now, what can you say about [itex]n^2- m^2[/itex]?
[itex] n^2 - m^2 = -1 [/itex] Which gives me a contradiction because this can only happen when n=0 and m =1, and the problem assumes n is strictly > 0.

Did I get it?

If so, then thank you!
 
  • #5
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Well, yeah, i think that should be right.

However, i would be inclined to proceede this way:

[tex]m^2-n^2=1=>(m-n)(m+n)=1[/tex] Then since both m, n are supposed to be integers(natural nrs) then the only possible way for this to be true is if:

[tex]m-n=m+n=1[/tex]

or

[tex]m-n=m+n=-1[/tex]

Because, we know that the only numbers that have inverses in Z are 1 and -1.

From the firs eq we get:m=1, n=0, which is a contradiction since the hypothesis says, n>0

From the second: m=-1 and n=0.

Which is again not possible.
 
  • #6
jgens
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Possibly another way of going about it:

We begin with some number n∈N, we may then claim n^2∈N. We now want to prove that √(n^2+1) ∉N. We do this by defining the next natural number m by m=(n+1); hence, m^2=(n+1)^2= n^2+2n+1. Therefore n^2<n^2+1<m^2 ∀n>0, proving that √(n^2+1) ∉N. Q.E.D.
 
Last edited:

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