Is the difference of a rational number and an irrational number always irrational? Explain.

Step-by-step explanation:

Yes, the irrational number stays irrational.

Make the assumption that this statement is false and that a rational and an irrational can produce a rational. x is irrational. Assume that a, b, c, and d are all integers.  If you like, they can be relatively prime to each other.

a/b + x = c/d                   Subtract a/b from both sides.

x = c/d - a/b                    Now put the 2 fractions over a common denominator.

x = (c*b - d*a)/(b*d)         Investigate what you have.

b*d are two integers multiplied together. multiplication of integers is closed (the answer will be an integer).

cb and da are also integers. subtracted integers are also closed under subtraction.

But

Here's the kicker. An irrational number is one that cannot be expressed as a fraction. Sqrt(2) for example cannot be shown to be rational. There is no fraction that equals sqrt(2)

So our proof is "bogus." It cannot work. An irrational number excludes the possibility of being able to be represented by a fraction by definition.

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.