Suppose "n" is positive integer number such that "n=n_1.n_2" where
"n_1 \\geq2 \\ and \\ n_2\\geq2."Therefore,"n_1\\ and \\ n_2" are two proper divisors other than 1.
If "n_1 \\ or \\ n_2" are equal to 1 then nothing to prove because 1 always fulfills that condition, i.e., "1\\leq n^\\frac{1}{2}" .
Claim":=" At least one of "n_1\\ or \\ n_2" is less than "n^{ \\frac{1}{2}}" .
Now we prove it using the method of contradiction.
On the contrary, assume that "n_1>n^\\frac{1}{2} \\ and \\ n_2>n^\\frac{1}{2}."
Then ,"n=n_1.n_2 >n^\\frac{1}{2}.n^\\frac{1}{2}"
"\\implies" "n>n"
Therefore, we get a contradiction. Hence an assumption was wrong.
Hence the claim was proved by the method of contradiction.
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