Question

3. Suppose e and f are real numbers where e>0 and f<0.

Answer 1

Part A:

Given that e and f are real numbers where e > 0 and f < 0.
Since e > 0, this means that e is positive and -e is negative.
Also, since f < 0, this means that f is negative.

The sum of two negative numbers gives a negative number.

Therefore, -e + f is negative.



Part B:

Given that e and f are real numbers where e > 0 and f < 0.
Since e > 0, this means that e is positive.
Also, since f < 0, this means that f is negative and -f is positive.

Recall that e - f = e + (-f)
The sum of two positive numbers gives a positive number.

Therefore, e - f is positive.



Part C:

Given that e and f are real numbers where e > 0 and f < 0.
Since e > 0, this means that e is positive and -e is negative.
Also, since f < 0, this means that f is negative.

Recall that f - e = f + (-e)
The sum of two negative numbers gives a negative number.

Therefore, f - e is negative.