Question

Use a proof of 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

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.