# I need some help :)!!

#### Eran

I am not sure I should post this on calculus forum, but I need help.

I want to prove that if x is a composite number and x <= n, where n is a
natural number,
there are i and j naturals, so that 1 < i <= [n/2]
([]-means floor value)

and 2 <= j <= [n/i], that making i*j = x.

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#### skipjack

Forum Staff
A composite number, x, is a positive integer that is divisible by a positive integer other than 1 and x.

If, for example, x = 2m, where m is a positive integer greater than 1,
the numbers i = 2 and j = m have the properties specified in the problem.

Does that provide enough help for you?

1 person

#### Eran

Yes, but how can I prove that? I can prove that there is an i that is between [n\2] and 1, that's no problem, but how can I prove that j can be between [n/i] and 2?

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#### skipjack

Forum Staff
Choose i to be the smallest divisor of x satisfying 1 < i $\small\leqslant$ [n/2], then choose j to be x/i. As i is at least 2, j is at most x/2.

As x $\small\leqslant$ n and j = x/i, j $\small\leqslant$ n/i, but you need to show that j $\small\leqslant$ [n/i].
What would x/i > [n/i] imply?

1 person

#### Eran

What would x/i > [n/i] imply?

it can imply that x/i>(n/i)-1 and because x<=n that imply that

n/i>n/i - 1 , so there isnt a contradiction, why?

#### skipjack

Forum Staff
Multiplying both sides of x/i > [n/i] by i gives x > [n/i]i,
but by the definition of [n/i], [n/i]i is the greatest multiple of i that doesn't exceed n;
it can't be exceeded by x, because x $\small\leqslant$ n.

1 person

#### Eran

[n/i]i is the greatest multiple of i that doesn't exceed n - I agree with that,
and also about that x>[n/i]i.

but that doesnt mean that x>n. so there isnt a contradiction between x>=n and x<n.

#### skipjack

Forum Staff
My word "it" referred to [n/i]i, not n. If [n/i]i were exceeded by x, x would be a greater multiple of i not exceeding n, which contradicts what you've already agreed with.

1 person

#### Eran

I am not sure I understand what you saying,
is there any problem that for x to be like that-> [n/i]i<=x<=n

#### skipjack

Forum Staff
The contradiction arises if $x > [n/i]i$, so $x \leqslant [n/i]i$.