Answer to Question #106146 in Real Analysis for lessinah

Question

Use a proof bt of the set of n1 and n2 is less than n^1/2y contradiction to establish the following:
If a positive whole number n can be expresses as n1 n2, where n1 is greater equals to 2 and n2 is greater equals to 2, then at least one element sets n1 and n2 is less than n^1/2

Answer 1

$n=n_1n_2,$ $n_1\geq2 , n_2\geq2$ .

Use a proof by contradiction. Let us assume that both of $n_1,\,n_2$ are greater than or equal to $n^{1/2}.$

Case I

$n_1=n^{1/2}$

$n_2=n^{1/2}$

$n_1n_2=n^{\frac12+\frac12}=n$

Case II

$n_1=n^{\frac12+x}; x>0$

$n_2=n^{\frac12+y}; y>0$

$n_1n_2=n^{\frac12+x}n^{\frac12+y}; x>0 ; y>0$

$n_1n_2=nn^{x+y}; x>0 ; y>0$

So, $x+y=0$

$x=-y$ but $x>0,y>0$

which is a contradiction.

So,If a positive whole number $n$ can be expressed as $n_1 n_2$ , where $n_1$ is greater than or equal to 2 and $n_2$ is greater than or equal to 2, then at least one element of $n_1$ and $n_2$ is less than $n^{1/2}$ .