Explain why sqrt(a+b) does not equal sqrt a + sqrt b, for a>0 and b>0?
because
(a+b)^2 ≠ a^2 + b^2
sqrt is not a linear operator
How about an indirect proof:
Assume that √(a+b) = √a + √b , for all a, b > 0
all we need is one counter-example to show my assumption is false:
is √(9 + 16) = √9 + √16 ???
LS = √25 = 5
RS = 3 + 4 = 7
RS ≠ LS
so I have a case where my assumption is false, all done
In mathematics, it is important to understand when certain equations or expressions are valid and when they are not. One such case is the equation sqrt(a+b) ≠ sqrt(a) + sqrt(b), where a and b are both greater than zero (a > 0 and b > 0).
To demonstrate why this equation is not true in general, let's consider a simple example. Let's take a = 1 and b = 1. According to the equation, sqrt(1 + 1) should be equal to sqrt(1) + sqrt(1).
Using the equation, we find that sqrt(2) equals sqrt(1) + sqrt(1). However, when we simplify the right side, we get sqrt(2) ≠ 1 + 1 or sqrt(2) ≠ 2. This is clearly not true since sqrt(2) is approximately 1.4142.
Therefore, we can conclude that sqrt(a+b) does not equal sqrt(a) + sqrt(b) when a and b are both greater than zero.
To understand why √(a + b) does not equal √a + √b, we can use a counterexample.
Let's pick values for a and b such that a = 4 and b = 1.
Now, let's evaluate both sides of the equation using these values:
√(a + b) = √(4 + 1) = √5 ≈ 2.24
√a + √b = √4 + √1 = 2 + 1 = 3
As you can see, the two sides of the equation yield different results.
To prove why they are not equal, we need to square both sides of the equation:
(√(a + b))^2 = (√5)^2 = 5
(√a + √b)^2 = (2 + 1)^2 = 3^2 = 9
Now, we can clearly see that 5 is not equal to 9.
Therefore, we can conclude that √(a + b) does not equal √a + √b in general.
In general, when we have two square roots added together, we cannot simply combine them as if they were non-radical terms. The properties of square roots do not allow us to directly add or subtract them without considering more advanced operations like using conjugates or square root equivalents.