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Thanks for the help for the previous proof. Now I am stuck on this statement I need to prove.

For any natural number $n$ there is an odd integer $m$ such that $n^2 < m < (n+1)^2$.

I got to: $$n^2 < 2n+1 < n^2 + 2n + 1$$

I am not sure how to proceed. Please help.

amWhy
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Brenda
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4 Answers4

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You can’t say that $n^2<2n+1<n^2+2n+1$: in general it’s not true that $n^2<2n+1$. What you need to do is show that there is at least one odd integer strictly between $n^2$ and $n^2+2n+1$.

HINT:

  • There are $2n$ integers $m$ such that $n^2<m<n^2+2n+1$; why?
  • If $n$ is a positive integer, what’s the smallest that $2n$ could possibly be?
  • How many integers in a row do you need to have in order to be certain that at least one of them is odd?
Brian M. Scott
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  • So, i subtract the right from the left and add 1 to get the number of numbers between them. n^2 + 2n + 1 - n^2 - 1 equals 2n. Since lowest possible n is 1, then there are at least 2 integers between them. When you have 2 consecutive integers, one must be negative. Is this correct? – Brenda Nov 11 '14 at 23:14
  • @Brenda: I think that you meant ‘and subtract $1$’ rather than ‘and add $1$’, and odd rather than negative. If so, you’re quite correct. – Brian M. Scott Nov 11 '14 at 23:15
  • I did. Sorry. Thank you so much, Brian. The hints really helped. – Brenda Nov 11 '14 at 23:18
  • @Brenda: No problem. You’re very welcome. – Brian M. Scott Nov 11 '14 at 23:19
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Hint: $(n+1)^2-n^2= 2n+1 \ge 3$ if $n\ge 1$.

lhf
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Suppose some number $(n+1)^2$ on the number line, where $n \in \mathbb{N}$. Evaluating that expression out, the number is equal to $n^2+2n+1.$ Then consider the number $n^2$. If we were to look at each number on the number line, the difference between both numbers would be: $$(n^2+2n+1)-n^2 = 2n+1.$$

How far are odd numbers away from each other on the number line? They are always two numbers apart. The $2n+1$ denotes how far the original two numbers, $n^2+2n+1$, and $n^2$, are apart from each other. As long as $n$ is a natural number, the two numbers will always be distance $3$ or more from each other.

And in that distance, there is bound to be an odd number, since odd numbers are exactly two numbers apart from each other.

daOnlyBG
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Proof. Induction on $n$. The base case $n=1$ is clearly. Now we suppose inductively true for $n$; we prove for $n+1$. So we need to find an odd integer between $(n+1)^2$ and $(n+2)^2$, i.e., between $n^2+2n+1$ and $n^2+4n+4$. Let $m$ be the odd integer such that $n^2<m<(n+1)^2$ by the inductive hipothesis. Since $n^2<m$, adding $2n+1$, we have $n^2+2n+1<m+2n+1$, i.e., $(n+1)^2<m+2n+1$; thus $(n+1)^2<m+2n+2$. Since $m<(n+1)^2=n^2+2n+1$, adding $2n+3$, we have $m+2n+3<n^2+4n+4$, i.e., $m+2n+3<(n+2)^2$; thus $m+2n+2<(n+2)^2$ Hence we have conclude that $$(n+1)^2<m+2n+2<(n+2)^2.$$ Since $m$ is odd, $2n$ is even and $2$ is even, we have $m+2n+2$ is odd, as desired. This close the induction.

Cristhian Gz
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