why a
(-)x(-)=(+)
(-)x(+)= (-)
2 answers
Those are the rules for multiplication of signed numbers.
If x>0 and a>0 and accepting that (+)(+) =(+) as well as the distributive property, lets investigate
x(a + (-a)), expanding we get
ax + (-a)x
we know that a + (-a) is zero, since the sum of two opposites is zero.
So the result above has to be zero.
Therefore (-a)x = -ax and a positive times a negative is negative.
now repeat the above steps with
(-x)(a + (-a)), again this result should be zero
expanding by the distributive property we get
(-x((a) + (-x)(-a)
we established that (-x)(a) is -ax, that is it is a negative number, so (-x)(-a) has to be its positive opposite to make the sum zero
so (-a)(-x) = +ax
I used to show the rules by relating them to playing a game in sports
Assume a good player is +
and a poor player is -
Assume playing is +
and sitting on the bench is -
1. A good player is playing, that's good +x+ = +
2. A poor player is playing, not good -x+ = -
3. A good player is on the bench, not good +x- = -
4. A poor player on the bench, that's good -x- = +
x(a + (-a)), expanding we get
ax + (-a)x
we know that a + (-a) is zero, since the sum of two opposites is zero.
So the result above has to be zero.
Therefore (-a)x = -ax and a positive times a negative is negative.
now repeat the above steps with
(-x)(a + (-a)), again this result should be zero
expanding by the distributive property we get
(-x((a) + (-x)(-a)
we established that (-x)(a) is -ax, that is it is a negative number, so (-x)(-a) has to be its positive opposite to make the sum zero
so (-a)(-x) = +ax
I used to show the rules by relating them to playing a game in sports
Assume a good player is +
and a poor player is -
Assume playing is +
and sitting on the bench is -
1. A good player is playing, that's good +x+ = +
2. A poor player is playing, not good -x+ = -
3. A good player is on the bench, not good +x- = -
4. A poor player on the bench, that's good -x- = +