Why can't the square root of a^2 + b^2 be simplified?

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Why can't the square root of a^2 + b^2 be simplified?

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Answer 1 · 2015-05-31T07:02:55

If we substitute a and b to equal 6 for example
it would be $\sqrt{6^{2} + 6^{2}}$ $sqrt(6^2+6^2)$ it would equal 8.5(1.d.p) as it would be written as $\sqrt{36 + 36}$ $sqrt(36+36)$ giving a standard form as $\sqrt{72}$ $sqrt72$

However if it was ${\sqrt{6}}^{2} + {\sqrt{6}}^{2}$ $sqrt6^2+sqrt6^2$ it would equal 12 as the $\sqrt{}$ $sqrt$ and $^2$ $^2$ would cancel out to give the equation 6+6

Therefore $\sqrt{a^{2} + b^{2}}$ $sqrt(a^2+b^2)$ cannot be simplified unless given a substitution for a and b.

I hope this isn't too confusing.

Answer 2 · 2015-05-31T09:30:37

Suppose we try to find a 'simpler' expression than $\sqrt{a^{2} + b^{2}}$ $sqrt(a^2+b^2)$

Such an expression would have to involve square roots or $n$ $n$ th roots or fractional exponents somewhere along the way.

Hayden's example of $\sqrt{6^{2} + 6^{2}}$ $sqrt(6^2+6^2)$ shows this, but let's go simpler:

If $a = 1$ $a=1$ and $b = 1$ $b=1$ then $\sqrt{a^{2} + b^{2}} = \sqrt{2}$ $sqrt(a^2+b^2) = sqrt(2)$

$\sqrt{2}$ $sqrt(2)$ is irrational. (Easy, but slightly lengthy to prove, so I won't here)

So if putting $a$ $a$ and $b$ $b$ into our simpler expression only involved addition, subtraction, multiplication and/or division of terms with rational coefficients then we would not be able to produce $\sqrt{2}$ $sqrt(2)$ .

Therefore any expression for $\sqrt{a^{2} + b^{2}}$ $sqrt(a^2+b^2)$ must involve something beyond addition, subtraction, multiplication and/or division of terms with rational coefficients. In my book that would be no simpler than the original expression.

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Why can't the square root of a^2 + b^2 be simplified?

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