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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First, the principle of Check and Balance applies. The Check and Balance system is devised to have all the branches of the government from overpowering one another. This is done through having independent powers in the government.
The next would be the principle that states that all powers not under the Federal government would be under the power of the State.
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Scatter plot, bar graph
Step-by-step explanation:
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Step-by-step explanation:
Incomplete question
This is the associative property of addition because the grouping of addends does not change the some no matter where you place the numbers in the parenthesis.
