Which operations is the following set closed under: {... , -3, -1, 1, 3, ...}?

2 answers:

6 0

I'd say it's closed for multiplication and division.
Any member of the set, when multiplied or divided by
any other one, produces another member of the set.

Ganezh [65]3 years ago

5 0

Not addition, since $-1+1=0$ and 0 doesn't belong to the set.

Not subtraction, since $1-1=0$ .

It is closed under multiplication. The set appears to contain all non-zero multiples of 3.

Not division, since dividing some of the numbers results in rationals which do not belong to the set.

What other operations are you supposed to consider?

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<img src="https://tex.z-dn.net/?f=6%2F5%20%28cos%28120%29%20%2B%20isin%28120%29%29%203%2F5%20%28cos%2845%29%20%2B%20i%20sin%2845

marysya [2.9K]

The product is equal to:

$\frac{18}{25}(cos(165) + i*sin(165))$

<h3>How to solve the product?</h3>

Remember that we can write a complex number in polar form as:

$R*e^{i*a} = R*(cos(a) + i*sin(a))$

Then the given product:

$\frac{6}{5}*(cos(120) + i*sin(120))*\frac{3}{5}*(cos(45) + i*sin(45))$

can be rewritten to:

$(\frac{6}{5}*e^{i*120})*(\frac{3}{5}*e^{i*45})$

Now is easier to solve the product:

$(\frac{6}{5}*e^{i*120})*(\frac{3}{5}*e^{i*45})\\\\= \frac{6}{5} *\frac{3}{5} *e^{i*(120 + 45)}\\\\= \frac{18}{25}*e^{i*165}\\\\= \frac{18}{25}(cos(165) + i*sin(165))$