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solve question 7、18、21 a mn is divisible by 15 Example 3 b in'h is diviville by ...
Jun 30, 2024
solve question 7、18、21
Solution by Steps
step 1
Let mm and nn be perfect cubes. This means m=a3m = a^3 and n=b3n = b^3 for some integers aa and bb
step 2
Consider the product mnmn. Substituting the values of mm and nn, we get mn=a3⋅b3mn = a^3 \cdot b^3
step 3
Using the property of exponents, we can rewrite a3⋅b3a^3 \cdot b^3 as (a⋅b)3(a \cdot b)^3
step 4
Since a⋅ba \cdot b is an integer, (a⋅b)3(a \cdot b)^3 is a perfect cube
Answer
The product of two perfect cubes is a perfect cube.
Key Concept
Perfect Cubes
Explanation
The product of two perfect cubes is always a perfect cube because the exponents add up to form another cube.
Question 18
step 1
The contrapositive of the statement "If 8n−18^n - 1 is prime, then nn is odd" is "If nn is even, then 8n−18^n - 1 is not prime"
step 2
Let nn be an even integer. Then n=2kn = 2k for some integer kk
step 3
Substitute n=2kn = 2k into 8n−18^n - 1: 82k−18^{2k} - 1
step 4
Using the difference of squares, 82k−1=(8k−1)(8k+1)8^{2k} - 1 = (8^k - 1)(8^k + 1)
step 5
Since 8k−18^k - 1 and 8k+18^k + 1 are both greater than 1 for k≥1k \geq 1, their product is not prime
Answer
If nn is even, then 8n−18^n - 1 is not prime.
Key Concept
Contrapositive
Explanation
The contrapositive of a statement is logically equivalent to the original statement. Proving the contrapositive proves the original statement.
Question 21
step 1
Define the function f(x)=x2x−1f(x) = \frac{x^2}{x-1}
step 2
Assume, for contradiction, that 11 is in the range of ff
step 3
This means there exists some xx such that f(x)=1f(x) = 1
step 4
Set f(x)=1f(x) = 1: x2x−1=1\frac{x^2}{x-1} = 1
step 5
Multiply both sides by x−1x-1: x2=x−1x^2 = x-1
step 6
Rearrange the equation: x2−x+1=0x^2 - x + 1 = 0
step 7
Solve the quadratic equation using the discriminant: Δ=b2−4ac=(−1)2−4(1)(1)=1−4=−3\Delta = b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3
step 8
Since the discriminant is negative, there are no real solutions to the equation
step 9
Therefore, 11 is not in the range of ff
Answer
11 does not belong to the range of f(x)=x2x−1f(x) = \frac{x^2}{x-1}.
Key Concept
Range of a Function
Explanation
By proving that the equation x2x−1=1\frac{x^2}{x-1} = 1 has no real solutions, we show that 11 is not in the range of the function.
Solution by Steps
step 1
Assume for contradiction that 2x−32x - 3 is rational
step 2
If 2x−32x - 3 is rational, then it can be written as ab\frac{a}{b} where a,b∈Za, b \in \mathbb{Z} and b≠0b \neq 0
step 3
Therefore, 2x−3=ab2x - 3 = \frac{a}{b}
step 4
Solving for xx, we get 2x=ab+32x = \frac{a}{b} + 3
step 5
Simplifying further, x=a2b+32x = \frac{a}{2b} + \frac{3}{2}
step 6
Since a,b∈Za, b \in \mathbb{Z}, a2b\frac{a}{2b} is rational and 32\frac{3}{2} is rational
step 7
The sum of two rational numbers is rational, so xx must be rational
step 8
This contradicts the assumption that xx is irrational
step 9
Therefore, our assumption that 2x−32x - 3 is rational must be false
step 10
Hence, if xx is irrational, then 2x−32x - 3 must be irrational
Answer
If xx is irrational, then 2x−32x - 3 is irrational.
Key Concept
Proof by Contradiction
Explanation
We assumed the opposite of what we wanted to prove and showed that it led to a contradiction, thus proving the original statement.
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