# Squaring both sides when units are different?

Given $((9) \text{inches})^{1/2} = ((0.25) \text{yards})^{1/2}$, then which of the following statements is true?

1. $((3) \text{inches}) = ((0.5) \text{yards})$
2. $((9) \text{inches}) = ((1.5) \text{yards})$
3. $((9) \text{inches}) = ((0.25) \text{yards})$
4. $((81) \text{inches}) = ((0.0625) \text{yards})$

My question is : Can I apply here as $x^{1/2}=y^{1/2}$ then square both sides $\implies x=y$. But as given units are different. So,

• Are you sure it's not $9^{1/2}\text{inches}=0.25^{1/2}\text{yards}$? Because I don't know what it would mean to take the square root of of an actual physical inch. – Jack M Feb 6 '16 at 12:12
• @JackM: For instance, $x=\frac12 t^2$ describes a body moving with constant acceleration of $1$ inch/second$^2$. For a given distance $x$, the time taken to travel $x$ inches is $\sqrt{2x}$ seconds. – TonyK Feb 6 '16 at 12:32
• @TonyK Right, but $x$ here refers to the amount of inches, not the inches themselves, so it's still $\sqrt{2x}\ \text{inches}$, not $\sqrt{2(x\text { inches})}$. – Jack M Feb 6 '16 at 13:08
• @JackM: Why can't you take the square root of an inch? It looks perfectly well-defined (if physically meaningless) to me. – Kevin Feb 6 '16 at 17:23
• @JackM There is actually a unit that uses the square root of length: fracture toughness has units of $psi \sqrt{inch}$. It's the only unit I know of off the top of my head that has units where one of the dimensions is raised to a non-integer power. – Kyle Feb 6 '16 at 17:54

If (one thing)$^\frac12$ = (another thing)$^\frac12$, then we can square both sides to get

one thing = another thing

Here, "one thing" is "$9$ inches", and "another thing" is "$0.25$ yards" (I don't know why you added all those parentheses!). Hence

$9$ inches = $0.25$ yards

Your worry about the units is irrelevant here, because they are inside the outer parentheses. This question is a reminder that $(9$ inches$)^\frac12$ is not the same as $9^\frac12$ inches.

• This is true when the things are numbers. But what is the square root of an inch, and how do you know that $\sqrt { 1inch} \sqrt { 1inch}= 1 inch$? I don't because I don't know what $\sqrt { 1 inch}$ is so I don't know whether it can be multiplied by itself. What is $\sqrt { 1 second}$? But Statement 3 is true anyway, by law,not by math. – DanielWainfleet Feb 6 '16 at 21:34
• @user254665: You should read all the answers and comments! – TonyK Feb 6 '16 at 21:40
• @user254665 Some of the issue is that mathematics does not define the concept of inches. It has numbers, and that's it. However, there is a concept known as a "physical quantity" in science, which does capture the idea of units such as inches, permitting science to use math to describe physical states. These physical quantities are codified as as the product of a number (such as 1) and a unit (such as inches). They have their own rules for manipulating these quantities, which have been chosen by scientists to lead to meaningful physical meanings. – Cort Ammon Feb 7 '16 at 3:58
• One of those rules is that if I have two numbers, a and b, and two units, X, and Y (so a*[X] and b*[Y] are both physical quantities, using brackets to make it clear this isn't normal multiplication), (a*[X])*(b*[Y]) = (ab)*[XY], So thus we can say sqrt(1)*[sqrt(inch)]*sqrt(1)*[sqrt(inch)] = 1*[sqrt(inch)*sqrt(inch)]. Then, a second rule of manipulating quantities states that if I have a unit [X], [X^a*X^b] = [X^(a+b)]. Thus, I can simplify further to 1*[inch]. Why can I do that? That's the accepted rules for multiplication for physical quantities in science. – Cort Ammon Feb 7 '16 at 4:01

If you have a square of side $9 \;\text{in}$, then the area of the square is $(9 \;\text{in})^2 = 81 \;\text{in}^2$. This is because when the sides of two squares are in the ratio $9:1$, their areas are in the ratio $81:1$, and we have defined the units $\text{in}^2$ so that $1\;\text{in}^2$ is the area of a square of side $1\;\text{in}$.

From this you might get the idea that units of measurement are just another algebraic quantity that is multiplied by the number on the left of the units. And indeed they do seem to work that way, because we have defined them so that they do. So if we have a quantity $ab$, with numeric value $a$ and units $b$, then when it make sense to square this quantity (such as when taking the area of a square of side $ab$), the result will be $(ab)^2 = a^2 b^2$.

This serves well when different units of the same dimension occur. For example, $3\;\text{inch} = 0.25\;\text{foot}$, so a square of side $3\;\text{inch}$ is also a square of side $0.25\;\text{foot}$, and its area is $9\;\text{inch}^2 = 0.0625\;\text{foot}^2$.

It appears that we can reverse this process by taking a square root, that is, $9\;\text{inch}^2 = 0.0625\;\text{foot}^2$ is the area of a square of side $3\;\text{inch} = 0.25\;\text{foot}$, so we might think that $(9\;\text{inch}^2)^{1/2} = 3\;\text{inch}$ and $(0.0625\;\text{foot}^2)^{1/2} = 0.25\;\text{foot}$.

If you suppose that it is possible for something to be measured in units of $\text{inch}^{1/2}$, and you continue to treat it as just another algebraic quantity, then it might appear to make sense to apply the formula $(ab)^{1/2} = a^{1/2} b^{1/2}$, so that $(9\;\text{inch})^{1/2} = 3\; \text{inch}^{1/2}$, and that you can therefore get back the original $9\;\text{inch}$ by squaring $(9\;\text{inch})^{1/2}$.

In that case, if $(9\;\text{inch})^{1/2}$ and $(0.25\;\text{yard})^{1/2}$ are two measurements of the exact same quantity, we should find that their squares also measure the same quantity as each other, so $9\;\text{inch} = 0.25\;\text{yard}$. And it happens that this last equation is true according to the real-life definitions of inch and yard.

But there is nothing (at least, nothing that you could reasonably be expected to know about) that is customarily measured in units of $\text{inch}^{1/2}$, so the idea that you can take the square root of $9\;\text{inch}$, including the units, is nonsensical in this setting. It is true in a formal sense, but that sense of the word "formal" means we manipulate the forms of things without thinking about what they are really supposed to mean.

In my opinion, the question is formal nonsense produced by someone who apparently gave little thought to its actual meaning. We see this sometimes in my country too (and even worse examples, where the thing you're asked to do is not merely silly, but wrong). The only advice I can give you is to play the game, manipulate the symbols formally even when the underlying idea is nonsense, and hope to play well enough that you will gain the privilege to study from people who can give you real insight into the subject they are teaching.

Update: From the comments on the question I have learned about fracture toughness, whose units involve a non-integer power of length (specifically, pressure times square root of length; there are units of length involved in the pressure as well, but they're integer powers so the result is a half-power no matter how you slice it). That's interesting news to me. In the first example of the use of this quantity in either of the places I've seen so far, the first thing we do is to square it, converting the half-power to a whole power, but presumably there is some other good reason to have the half-power in the first place.

There are a quite a few other places where we take the square root of something that has units of measurement, for example, in the formula for how long it takes to fall distance $y$ under gravitational acceleration $g$, $t = \sqrt{2y/g}$. There if you double the distance, the time increases by only $\sqrt2$. As far as the units are concerned, however, the length dimension of $y$ cancels the length dimension of $g$, leaving only a dimension of $\text{time}^2$ under the radical; and we are quite used to taking the square root of squared units to get back the original units. If it were customary to publish the value of $\sqrt{2/g}$ in reference books, rather than just $g$, we would have a constant of dimension $\text{time}/\text{length}^{1/2}$ to be multiplied by something of dimension $\text{length}^{1/2}$. I've never seen this, but it's possible there is some field of study I have not experienced yet where it is done that way.

• I think you are wrong to be so dismissive of this question. See the comment thread under the OP for why. – TonyK Feb 6 '16 at 13:56
• Go and read Kyle's comment under the OP (and then apologise to the question setter!). – TonyK Feb 6 '16 at 19:44
• @TonyK I did a little reading up on fracture toughness. Interesting stuff. I don't think this is what the question setter had in mind, and I really doubt that if they did it would be appropriate for this exam, but I've edited the answer to tone down my opinion about the question (though the question setter is very unlikely to see anything I write in any case) as well as to acknowledge applications of fractional powers of units that have been mentioned elsewhere. – David K Feb 6 '16 at 21:04
• As another example, the units of a quantum mechanical electronic wavefunction $\Psi\!\left(\vec r\right)$ in $\mathbb R^3$ are $\mathrm{m}^{-{3\over 2}}$, since $\rho\!\left(\vec r\right) = \Psi\!^*\!\left(\vec r\right)\Psi\!\left(\vec r\right)$ and $\int{\rho\!\left(\vec r\right)\,d\vec r} = 1$. – hBy2Py Feb 7 '16 at 4:02

Hint: Given $\sqrt{9x}=\sqrt{0.25y}\\ \Longrightarrow3\sqrt{x}=0.5\sqrt{y}\\ 3\sqrt{x}=\frac{\sqrt y}{2}\\ \sqrt y=6 \sqrt x\\ y=36 x$

Spoiler:

3. $((9) \text{inches}) = ((0.25) \text{yards})$

If you treat $\operatorname{in}$ and $\operatorname{yd}$ as if they were variables, then you can simplify

$\left( 9 \; \operatorname{in} \right)^{1/2} = \left( \dfrac 14 \; \operatorname{yd} \right)^{1/2}$

to

$3 \operatorname{in}^{1/2} = \dfrac 12 \; \operatorname{yd}^{1/2}$

then you can square both sides and get

$9 \; \operatorname{in} = \dfrac 14 \; \operatorname{yd}$

which is true. But I have no idea what $\operatorname{in^{1/2}}$ means physically. In most cases, you can treat units as if they were variables but there needs to be a physical interpretation of the units that you end up with or just becomes formal gibberish.

• Thanks for nice explanation. – ً ً Feb 7 '16 at 5:53

Your original equation is actually: 9 inches = .25 yards * (36 inches/yard)

(The unit conversion is implicit. When you make it explicit, it makes the units on both sides work).

When you take the square root of both sides, you get: 3 inches^.5 = .5 yards^.5 * (6 inches^.5/yards^.5)

This, while working, is kind of a meaningless equation, since the square root of an inch has no real physical meaning.

However, if you square both sides you get: 81 inches^2 = (1/16)yards^2 * (1296 inches^2/yards^2)

Since square inches and square yards have a physical representation (area), the equation not only makes sense, but gives you the conversion rate between square inches and square yards.

• Thanks for nice explanation. – ً ً Feb 7 '16 at 5:53

If you say for linear measure

$$1 ft = 12 \, in,$$ then for area measure $$1 ft^2 = 144 \, in^2,$$ and $$1 ft^3 = 1728 \, in^3,$$

For derived units

Speed

$$1 ft/ \min = 12/60 = 0.2 in/\sec,$$

Rate of discharge or volume rate

$$3600 m^3 / hour = 1 m^3/\sec,$$

and so on.

• Thanks for nice explanation. – ً ً Feb 7 '16 at 5:52