Explain why the product of a non-0 rational number and an irrational number is irrational?

An irrational number is a number that can not be written as the quotient of two integer numbers.

Then if we have:

A = a rational number

B = a irrational number.

Then we can write:

A = x/y

Then the product of A and B can be written as:

A*B = (x/y)*B

Now, let's assume that this product is a rational number, then the product can be written as the quotient between two integer numbers.

(x/y)*B = (m/n)

If we isolate B, we get:

B = (m/n)*(y/x)

We can rewrite this as:

B = (m*y)/(n*x)

Where m, n, y, and x are integer numbers, then:

m*y is an integer

n*x is an integer.

Then B can be written as the quotient of two integer numbers, but this contradicts the initial hypothesis where we assumed that B was an irrational number.

Then the product of an irrational number and a rational number different than zero is always an irrational number.

We need to add the fact that the rational number is different than zero because if:

B is an irrational number

And we multiply it by zero, we get:

B*0 = 0

Then the product of an irrational number and zero is zero, which is a rational number.

1     sepan

2     encuentren

3     tenga

4     hable

5   parezca

6   llegue

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.