The cube root of 2 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification. Assume the cube root of 2 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the cube root of 2, its cube must equal 2. That is, (a/b)3 = 2 a3/b3 = 2 a3 = 2b3. The right side is even, so the left side must be even also, thatis, a3 is even. Since a3 is even, a is also even (because the cube of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c. Now, (2c)3 = 2b3 8c3 = 2b3 4c3 = b3. The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b3 cannot be odd; it must be even. Therefore b is even as well. Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must <span>be false, and the cube root of 2 is irrational.
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Answer:
x = -36
Step-by-step explanation:
Answer:
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b) (-3 + 5i) (1) = -3 + 5i
Step-by-step explanation:
Answer: x = -15
Step-by-step explanation:
Think PEMDAS
First, we factor the parenthesis:
-4 (x-4) = -4(x) + -4(-4)
now we are left with -4x + 16 = -44
Next is to isolate the "x":
to do this, we must subtract 16 from -44
-44 - 16 = 60
now we have -4x = 60
Now we divide both sides by -4 in order to isolate "x"
dividing -4x/-4 = x
diving 60/-4 = -15
the answer is x= -15
