You can find counterexamples to disprove this claim. We have positive integers that are perfect square numbers; when we take the square root of those numbers, we get an integer.
For example, the square root of 1 is 1, which is an integer. So if y = 1, then the denominator becomes an integer and thus we get a quotient of two integers (since x is also defined to be an integer), the definition of a rational number.
Example: x = 2, y = 1 ends up with
which is rational. This goes against the claim that
is always irrational for positive integers x and y.
Any integer y that is a perfect square will work to disprove this claim, e.g. y = 1, y = 4, y= 9, y = 16. So it is not always irrational.
Answer
(a) 
(b) 
Step-by-step explanation:
(a)
δ(t)
where δ(t) = unit impulse function
The Laplace transform of function f(t) is given as:

where a = ∞
=> 
where d(t) = δ(t)
=> 
Integrating, we have:
=> 
Inputting the boundary conditions t = a = ∞, t = 0:

(b) 
The Laplace transform of function f(t) is given as:



Integrating, we have:
![F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.](https://tex.z-dn.net/?f=F%28s%29%20%3D%20%5B%5Cfrac%7B-e%5E%7B-%28s%20%2B%201%29t%7D%7D%20%7Bs%20%2B%201%7D%20-%20%5Cfrac%7B4e%5E%7B-%28s%20%2B%204%29%7D%7D%7Bs%20%2B%204%7D%20-%20%5Cfrac%7B%283%28s%20%2B%201%29t%20%2B%201%29e%5E%7B-3%28s%20%2B%201%29t%7D%29%7D%7B9%28s%20%2B%201%29%5E2%7D%5D%20%5Cleft%20%5C%7B%20%7B%7Ba%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
Inputting the boundary condition, t = a = ∞, t = 0:

Answer:
105
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
Given
-
[ -
c + 6c + 1 ] - 3c
Distribute the square bracket by - 
=
c - 3c -
- 3c ← combine like terms
=
c - 6c - 
=
c -
c - 
= -
c - 