Is the square root of 5 plus the square root 5 equal to the square root of 10?

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Is the square root of 5 plus the square root 5 equal to the square root of 10?

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Answer 1 · 2015-09-09T20:58:53

No:

$\sqrt{5} + \sqrt{5} = 2 \sqrt{5} = \sqrt{4} \sqrt{5} = \sqrt{20} \approx 4.472135955$ $sqrt(5)+sqrt(5) = 2sqrt(5) = sqrt(4)sqrt(5) = sqrt(20) ~~ 4.472135955$

$\sqrt{10} = \sqrt{2 \cdot 5} = \sqrt{2} \sqrt{5} \approx 3.16227766$ $sqrt(10) = sqrt(2*5) = sqrt(2)sqrt(5) ~~ 3.16227766$

Explanation:

In the above answer, I use $\sqrt{a b} = \sqrt{a} \sqrt{b}$ $sqrt(ab) = sqrt(a)sqrt(b)$ - which is true for any $a, b \geq 0$ $a, b >= 0$ .

Generalisation

Suppose $f$ $f$ is a function that takes any Real number and gives us another Real number.

Under what circumstances would we expect the following:

$f (a + b) = f (a) + f (b)$ $f(a+b) = f(a) + f(b)$ for all $a$ $a$ and $b$ $b$ ?

Actually the only kind of functions that behave like this are all linear functions which take the form:

$f (x) = m x + c$ $f(x) = mx + c$ , where $m$ $m$ and $c$ $c$ are constants.

In addition $f (0) = 0$ $f(0) = 0$ , so $c = 0$ $c = 0$ and $f (x)$ $f(x)$ must take the form:

$f (x) = m x$ $f(x) = mx$ for some constant $m$ $m$ .

The function $f (x) = \sqrt{x}$ $f(x) = sqrt(x)$ is not linear, so does not behave in this way.

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Is the square root of 5 plus the square root 5 equal to the square root of 10?

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