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Example: Let's presume one was attempting to isolate m below:

A common mistake would be: $k^2 = m^2 + n^2 \to k = m +n$

Even though: $k^2 = m^2 + n^2 \to k \neq m +n$

If you apply a square root to both sides of the equation, you will have an inequality.

Why is this true?

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    $\begingroup$ There is no inequality. Applying the square root gives $k = \pm \sqrt{m^2 + n^2}$, which is the correct way to apply the square root. The important part is to notice that $\sqrt{m^2 + n^2}$ is not the same as $\sqrt{m^2} + \sqrt{n^2}$ $\endgroup$ – wgrenard Mar 29 '16 at 0:50
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    $\begingroup$ What we need here is a cure for the “law of universal linearity”. $\endgroup$ – Rahul Mar 29 '16 at 1:07
  • $\begingroup$ To me it doesn't make much sense to ask "why" X doesn't equal Y when there's no reason to suggest they're equal in the first place. $\endgroup$ – arctic tern Mar 29 '16 at 2:50
  • $\begingroup$ The law does hold if the domain of the variables is restricted to the integers modulo 2. $\endgroup$ – Jan Stout Mar 29 '16 at 6:06
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If you put the question in a broader context you may understand it better.

Suppose $f$ is some function you can apply to numbers. It might be squaring, or square-rooting, or inverting, or raising to some other power, or taking logarithms or $\sin$ or $\cos$ or just adding 15. In none of these cases is $f(x+y)$ the same as $f(x) + f(y)$ (you should check). In general, you would not expect that coincidence.

The special case in which it is true is the function "multiply by a fixed quantity". That's the distributive law:

$$ c \times (x + y ) = c \times x + c \times y . $$

Many of the most common errors students make in algebra or precalculus come from thinking that those other functions behave this way too.

Edit: Just in case you missed @Rahul 's comment: What we need here is a cure for the “law of universal linearity”: Pedagogy: How to cure students of the "law of universal linearity"?

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  • $\begingroup$ Explained with a simple analogy, thank you. $\endgroup$ – MacroGuy Mar 29 '16 at 1:07
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    $\begingroup$ If you like the answer you can accept it, and upvote. You can upvote other answers too. $\endgroup$ – Ethan Bolker Mar 29 '16 at 1:09
  • $\begingroup$ A small point: there exist non-linear functions that satisfy $$f(x+y) = f(x)+f(y)$$ for all $x,y$. This was a bit unintuitive for me. Edit: It is possible to prove that these functions exist, but probably not possible to write one down explicitly ... very strange! $\endgroup$ – Zubin Mukerjee Mar 29 '16 at 1:56
  • $\begingroup$ @ZubinMukerjee True, but I won't edit "The special case" to "A special case" since the question is tagged algebra-precalculus. $\endgroup$ – Ethan Bolker Mar 29 '16 at 2:00
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Well this can be shown by the triangle inequality. Another way to look at it is for $k^2 = m^2 + n^2$ and $k = m + n$ you have that $(m+n)^2 = m^2 + n^2$ which is only true when either m or n are 0. In other words, the inequality only holds if you assume $m, n$ are nonzero.

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"Why is this true?" (that $f(x+y) \ne f(x)+f(y)$).

Because the functions that satisfy $f(x+y)=f(x)+f(y)$ and are not badly behaved (continuous will work) are all linear, so that $f(x) = cx$ for some real $c$.

So if you try $f(x) = \sqrt{x}$, you can not have $f(x+y) = f(x)+f(y)$.

There are (at least) two ways to prove this.

First, if $f(x+y) = f(x)+f(y)$, then $\sqrt{x+y} = \sqrt{x}+\sqrt{y}$, then, squaring, we get $x+y = x+y+2\sqrt{xy}$, so that $2\sqrt{xy} = 0$, so that $xy = 0$, or at least one of $x$ and $y$ is zero.

Second, if we use the result that $f(x) = cx$ for some real $c$, then $\sqrt{x} = cx$. Squaring and dividing by $x$, we get $1=c^2x$, which can not hold for any constant $c$.

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Consider $k,m,n$ as the sides of a triangle. Then:

  • $k^2 = m^2 + n^2$ means that the triangle is a right triangle.

  • $k = m +n$ means that the triangle is a degenerate triangle.

Clearly, no right triangle can be degenerate!

(Unless one of the sides is zero, in which case, yes, $k^2 = m^2$ implies $k=m$.)

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  • $\begingroup$ You can have degenerate right triangles if one side is zero, right? $\endgroup$ – Zubin Mukerjee Mar 29 '16 at 2:07
  • $\begingroup$ @ZubinMukerjee, right. I've edited my answer. $\endgroup$ – lhf Mar 29 '16 at 10:29
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Otherwise $\sqrt n = n$ for all natural number $n$.

More generally $\sqrt q = q$ for all positive rational number $q$.

Homework:

Is it true that: $\sqrt x = x$ for all positive real number $x$?

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