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vaieri [72.5K]
3 years ago
12

Use the unique factorization of integers theorem to prove the following statement. If n is any positive integer that is not a pe

rfect square, then ✓n is irrational. The following is a proposed proof by contradiction of the statement with at least one incorrect step. 1. Suppose not. Suppose there exists an integer n such that n is not a perfect square and n is rational. 2. By definition of rational number, there exist integers a and b such that - and b0. 3. Without loss of generality, we may assume that a and b have no common divisors. Squaring both sides of the equation gives n = 4, and multiplying by b gives nb = a. 4. By the unique factorization of integers theorem, a, b, and n have representations as products of primes that are unique except for the order in which the prime factors are written down. 5. Since n is not a perfect square, there is a prime factor p in n that occurs an odd number of times. 6. Since nb = a, then p is a factor of a. 7. Thus a and b share the common divisor p. 8. This contradicts the fact that a and b are relatively prime. 9. Hence the supposition is false and the given statement is true. Identify the mistake(s) in the proof. (Select all that apply.) U Statement 7 does not follow from statements 3, 4, 5, and 6. M Statement 6 does not follow from statements 3, 4, and 5. The proof assumes a and b have no common factors, but the statement should apply for all integers a and b. The assumption in Statement 1 should be that n is a perfect square and n is rational. The equation in Statement 2 is not justified by the definition of rational number. The proof excludes the case where b = 0, but the statement should apply for every integer b.
Mathematics
1 answer:
anyanavicka [17]3 years ago
3 0

Answer:

no

Step-by-step explanation:

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3 years ago
Assume there are 365 days in a year.
Dahasolnce [82]
The chance of student 1's birthday being individual is 365/365 or 100%.
Then the chance of student 2's birthday being different is 364/365.
Then it's narrowed down to 363/365 for student 3 and so on until you get all 10 students.
If you multiply all these values together, the probability would come out at around 0.88305182223 or 0.88.

To get all the same birthday you'd have to the chance of one birthday, 1/365 and multiply this by itself 10 times. This will produce a very tiny number. In standard form this would be 2.3827x10'-26 or in normal terms: 0.23827109210000000000000000, so very small.
4 0
3 years ago
Write 3x^2-18x-6 in vertex form
Vedmedyk [2.9K]
The standard form of a quadratic equation is \displaystyle{ y=ax^2+bx+c, while the vertex form is:

                      y=a(x-h)^2+k, where (h, k) is the vertex of the parabola.

What we want is to write \displaystyle{ y=3x^2-18x-6 as y=a(x-h)^2+k

First, we note that all the three terms have a factor of 3, so we factorize it and write:

\displaystyle{ y=3(x^2-6x-2).


Second, we notice that x^2-6x are the terms produced by (x-3)^2=x^2-6x+9, without the 9. So we can write:

x^2-6x=(x-3)^2-9, and substituting in \displaystyle{ y=3(x^2-6x-2) we have:

\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11].

Finally, distributing 3 over the two terms in the brackets we have:

y=3[x-3]^2-33.


Answer: y=3(x-3)^2-33
6 0
3 years ago
Which statements are correct about the two way frequency table? Check all that apply. The table displays quantitative data, the
Andreyy89

Answer:

A the table display qualitative data, C two teachers do not wear glasses, E a total of 6 teachers were polled

7 0
3 years ago
Read 2 more answers
A boat costs $12,250 & decreases in value by 11% per year how much will the boat be worth after 4 years
kirill115 [55]
So it decreases by 11% which means by the next year it's worth 89% of the original price, write 89% like a decimal. then put it to the power of 4 (4 years) and multiply by the original number
.89 ^ 4 * 12250 = 7685.92
5 0
3 years ago
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