Mistakes give rise to Problems- 4

The property is false when $\text{arg}(x)\ne \text{arg}(y)$ .

Since we are working in the reals, this happens when either $x$ is positive and $y$ is negative or $x$ is negative and $y$ is positive.

There are $10$ negative numbers to choose from and $10$ positive ones to choose from; thus, there is $100$ ways to make the property false in each case giving $200$ ways to make the property false.

There are $21$ numbers total so there are $21^2=441$ total different ways to pick $x$ and $y$ .

Thus, our answer is $441-200=\boxed{241}$ .

When a>0, b >= 0.........10 * 11 = 110 choices.
When a<0, b <= 0.........10 * 11 = 110 choices.
When a = 0, b has all.........21 = 21 choices....Total 241.
Daniel Liu probably has better shorter approach.

In order for the false property to be true, a number will have to be paired with a number of the same sign or with 0. First, we count the pairs including positive numbers or 0.

$11 \times 11 = 121$

Next, we count the pairs including negative numbers or 0.

$11 \times 11 = 121$

Now we sum the two.

$121 + 121 = 242$

However, we must recognize that we have counted the ordered pair $(0,0)$ twice, so we must subtract $1$ .

$242 - 1 = \boxed{241}$

2 (11)^2 - 1 = 241

$21^2 - 10^2 - 10^2 = 241$

|a|+|b|=|a+b|

This is only possible if

1) Both a and b are positive

For a=0,1,2......10
b=0,1,2....10 therefore 121 possibilities

2) Both a and b are negative

Therefore possibilities=121 but since we have already counted a=0,b=0 once

Thus the total number of possibilities= 121+121-1=241