Algebra+ 06
Problem A
Prove that sum of square roots of 2, 3 and 5 is an irrational number.
Hint A
Assume, this sum is rational, that is
√2 + √3 + √5 = p/q
where p and q are integer numbers without common divisors (if they do, we can reduce the fraction by dividing a numerator p and denominator q by a common divisor without changing the value of a fraction).
Then simplify the above expression by getting rid of square roots and prove that p must be an even number and, therefore, can be represented as p=2r.
Then prove that q must be even as well, and, therefore, p and q have a common divisor 2, which we assumed they don't.
Problem B
Given a polynomial represented as a product
P = (5−4x)1000·(3x−4)1001
Assume, we open all the parenthesis and combine all similar terms to express this polynomial in a canonical form
P = Σn∈[0,2001]An·x2001−n
What would be a sum of all the coefficients An?
Hint B:
Do not attempt to use Newton's binomial and find the answer by explicitly performing all the operations to convert the given expression into canonical polynomial form.
There is a better and very quick way.
Answer B:
Sum of all the coefficients will be equal to −1.
Problem C
Find all prime integer x and y, for which the following is true
13·(x + y) = 3·(x² − x·y + y²)
Hint C
Express the given equation as a quadratic equation for x with coefficients as functions of y.
Then, to have solutions for x, a discriminant must be non-negative, which reduces the possible values for y to be in an interval (0,10).
Answer C
Only a pair of numbers 2 and
So, the solutions are (x=2,y=7) or (x=7,y=2).
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