- #1

Ackbach

Gold Member

MHB

- 4,155

- 92

In my experience teaching Calculus at the university level, there were a number of algebra mistakes I would see quite often. In an effort to help you, the ordinary math student, I would like to create a post here that outlines a few of these mistakes, and how to avoid them. The general format is going to be: the mistake, why it's a mistake, followed by how to remember not to make that mistake.

In addition, there are a few techniques that keep coming up over and over again, that students need to have in their toolbox. I'll list a few of those for you as well.

But first, the mistakes.

This one is also known as the "Freshman's Dream". It looks like it ought to be true, but it simply isn't. Why isn't it true? Because you've left out the cross terms; it's also not true because the squaring function is not linear. The LHS of the above

$$(x+y)^{2}=(x+y)(x+y)=x^{2}+xy+yx+y^{2}=x^{2}+xy+xy+y^{2}=x^{2}+2xy+y^{2}.$$

And now, you can see that unless the cross term is zero (not true in general), the Freshman's Dream is not true.

How can you remember this? Probably the easiest way is to

Why isn't this true? Because the square root function is not linear. You can test it out with some Pythagorean triples. Suppose $a=9, b=16$. Then you've got

$$5=\sqrt{25}=\sqrt{9+16}\overset{?}{=} \sqrt{9}+\sqrt{16}=3+4=7.$$

Well, $5\not=7$, so evidently this one doesn't work, either. So, in this case, the reason why it doesn't work, and the memory device for remembering that it doesn't work can be the same thing.

$$\sin(x+y)\not=\sin(x)+\sin(y),\quad \ln(x+y)\not=\ln(x)+\ln(y),\quad e^{x+y}\not=e^{x}+e^{y},\dots$$

So, for the first, there's a formula you should know:

$$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y).$$

If you're familiar with matrix rotations, there's an elegant way to remember this formula, along with the cosine version:

$$\begin{bmatrix}\cos(x+y)\\ \sin(x+y)\end{bmatrix}=\begin{bmatrix}\cos(x) &-\sin(x)\\ \sin(x) &\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)\\ \sin(y)\end{bmatrix}$$

For the second inequation, there's no way to simplify the LHS. Instead, what you're probably thinking is that

$$\ln(xy)=\ln(x)+\ln(y),$$

which is true. Check this equation with $x=1$ and $y=e$. You get $1=\ln(e)=\ln(1\cdot e)=\ln(1)+\ln(e)=0+1$. It doesn't work with the inequation above. Brush up on your logarithms!

For the last one, it's similar to the logarithm, only inverted:

$$e^{x+y}=e^{x}e^{y}.$$

Here, check it out with $x=0$ and $y=1$. You're probably seeing a pattern here: if you're unsure of an equation, check it out with specific numbers! Don't use completely trivial numbers like $0$, unless it's warranted. You might get a fluke where your incorrect equation happens to work with a special case. Check it out with multiple cases.

For the same reasons as before, this simply doesn't work. Most functions are not linear, and so you can't expect multiplication to distribute over most functions. When in doubt, don't! Or check it with specific numbers.

This one's a shocker for many people. Why can't I cancel? Well, you can, but only if $x\not=-1$. So you can do this:

$$\frac{x(x+1)}{x+1}=x,\quad\text{for}\;x\not=-1.$$

Why is this the case? Because with the initial inequation I mentioned, you're really claiming that the

Do I mean $\sin(x)+y$ or do I mean $\sin(x+y)$? I once worked a problem from a physics textbook where the authors neglected to include parentheses on a rather complicated argument to the $\sin$ function. I got the answer wrong, because I assumed the incorrect version. Ever since then, I've been a stickler for including parentheses on function arguments. Don't write so that you can be understood! Write so that you

So here are a few mistakes. What are some things I recommend doing?

This technique is invaluable. I even used it in my Ph.D. dissertation. You can use it to simplify expressions, maximize or minimize expressions, find the centers of circles, find information on various other conic sections, etc. Know how to solve the general quadratic in your sleep using this technique:

$$ax^{2}+bx+c=0,\quad a\not=0.$$

$$x^{2}+\frac{b}{a}\,x+\frac{c}{a}=0$$

$$x^{2}+\frac{b}{a}\,x+\frac{b^{2}}{4a^{2}}-\frac{b^{2}}{4a^{2}}+\frac{4ac}{4a^{2}}=0$$

$$\left(x+\frac{b}{2a}\right)^{2}=\frac{b^{2}-4ac}{4a^{2}}$$

$$x+\frac{b}{2a}=\pm\frac{\sqrt{b^{2}-4ac}}{2a}$$

$$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$

$$x=\frac{2}{y}\implies y=\frac{2}{x}.$$

You can swap the LHS and the denominator of the RHS in this fashion, provided you don't then divide by zero (remember the domain issue I mentioned above!)

$$2-x=y\implies 2-y=x.$$

Similar to above.

$$\frac{\sqrt{1+h}-\sqrt{1}}{h}=\frac{\sqrt{1+h}-\sqrt{1}}{h}\cdot\frac{\sqrt{1+h}+\sqrt{1}}{\sqrt{1+h}+\sqrt{1}}=\frac{1+h-1}{h(\sqrt{1+h}+\sqrt{1})}=\frac{1}{\sqrt{1+h}+ \sqrt{1}},$$

for $h\not=0$.

This is useful in calculus when evaluating limits.

These are all that come to mind at the moment. If you are aware of any other common mistakes or valid and useful techniques you see, please reply to this thread. I will moderate this thread, but you can help make this thread more valuable with your experience. Thank you!

Questions and comments should be posted here:

http://www.mathhelpboards.com/f49/commentary-algebra-dos-donts-4224/#post19141

In addition, there are a few techniques that keep coming up over and over again, that students need to have in their toolbox. I'll list a few of those for you as well.

But first, the mistakes.

**Mistake # 1: Distributing addition over squaring:**$(x+y)^{2}\not=x^{2}+y^{2}$.This one is also known as the "Freshman's Dream". It looks like it ought to be true, but it simply isn't. Why isn't it true? Because you've left out the cross terms; it's also not true because the squaring function is not linear. The LHS of the above

**in**equation needs multiplying out term-by-term thus:$$(x+y)^{2}=(x+y)(x+y)=x^{2}+xy+yx+y^{2}=x^{2}+xy+xy+y^{2}=x^{2}+2xy+y^{2}.$$

And now, you can see that unless the cross term is zero (not true in general), the Freshman's Dream is not true.

How can you remember this? Probably the easiest way is to

**plug in specific numbers**and see if it works. Try $x=1$ and $y=2$. Then the LHS is $9$, and the RHS is $5$. Since they're not equal, the Freshman's Dream is incorrect.**$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$.**

Mistake # 2: Distributing addition over the square root:Mistake # 2: Distributing addition over the square root:

Why isn't this true? Because the square root function is not linear. You can test it out with some Pythagorean triples. Suppose $a=9, b=16$. Then you've got

$$5=\sqrt{25}=\sqrt{9+16}\overset{?}{=} \sqrt{9}+\sqrt{16}=3+4=7.$$

Well, $5\not=7$, so evidently this one doesn't work, either. So, in this case, the reason why it doesn't work, and the memory device for remembering that it doesn't work can be the same thing.

Mistake # 3: Distributing addition over pretty much any function:Mistake # 3: Distributing addition over pretty much any function:

$$\sin(x+y)\not=\sin(x)+\sin(y),\quad \ln(x+y)\not=\ln(x)+\ln(y),\quad e^{x+y}\not=e^{x}+e^{y},\dots$$

So, for the first, there's a formula you should know:

$$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y).$$

If you're familiar with matrix rotations, there's an elegant way to remember this formula, along with the cosine version:

$$\begin{bmatrix}\cos(x+y)\\ \sin(x+y)\end{bmatrix}=\begin{bmatrix}\cos(x) &-\sin(x)\\ \sin(x) &\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)\\ \sin(y)\end{bmatrix}$$

For the second inequation, there's no way to simplify the LHS. Instead, what you're probably thinking is that

$$\ln(xy)=\ln(x)+\ln(y),$$

which is true. Check this equation with $x=1$ and $y=e$. You get $1=\ln(e)=\ln(1\cdot e)=\ln(1)+\ln(e)=0+1$. It doesn't work with the inequation above. Brush up on your logarithms!

For the last one, it's similar to the logarithm, only inverted:

$$e^{x+y}=e^{x}e^{y}.$$

Here, check it out with $x=0$ and $y=1$. You're probably seeing a pattern here: if you're unsure of an equation, check it out with specific numbers! Don't use completely trivial numbers like $0$, unless it's warranted. You might get a fluke where your incorrect equation happens to work with a special case. Check it out with multiple cases.

**Mistake # 4: Distributing multiplication or division over any of these things.**For the same reasons as before, this simply doesn't work. Most functions are not linear, and so you can't expect multiplication to distribute over most functions. When in doubt, don't! Or check it with specific numbers.

**Mistake # 5: Canceling factors without provisos.**$\frac{x(x+1)}{x+1}\not=x$.This one's a shocker for many people. Why can't I cancel? Well, you can, but only if $x\not=-1$. So you can do this:

$$\frac{x(x+1)}{x+1}=x,\quad\text{for}\;x\not=-1.$$

Why is this the case? Because with the initial inequation I mentioned, you're really claiming that the

**function**on the LHS is equal to the**function**on the RHS. But two functions are only equal if their domains are the same, and their rules of association are the same. The domain of the function on the left is $\mathbb{R}\setminus\{-1\}$, and the domain of the function on the right is $\mathbb{R}$. Since the domains are not the same, you cannot**just**cancel. Now, if you happen to know that in your application, $x$ is never equal to $-1$, then by all means, cancel! One notable example of this is if you're computing a limit in calculus when $x\to-1$. In this limit, you never let $x=-1$, so you're fine. You just have to be careful in equating two functions to make sure the domains are the same, or that you don't care about those regions where the domains are not the same. The moral of the story is: be intentional about your canceling, and think through these issues!**Mistake # 6: Omitting parentheses in function arguments.**$\sin x+y=?$.Do I mean $\sin(x)+y$ or do I mean $\sin(x+y)$? I once worked a problem from a physics textbook where the authors neglected to include parentheses on a rather complicated argument to the $\sin$ function. I got the answer wrong, because I assumed the incorrect version. Ever since then, I've been a stickler for including parentheses on function arguments. Don't write so that you can be understood! Write so that you

*can't*be*mis*understood! Yes, it might take a tad longer initially. However, the time you save in being clearly understood by others will often more than make up for that "lost" time.So here are a few mistakes. What are some things I recommend doing?

**Do # 1: Completing the square.**$x^{2}+4x+9=x^{2}+4x+4+5=(x+2)^{2}+5.$This technique is invaluable. I even used it in my Ph.D. dissertation. You can use it to simplify expressions, maximize or minimize expressions, find the centers of circles, find information on various other conic sections, etc. Know how to solve the general quadratic in your sleep using this technique:

$$ax^{2}+bx+c=0,\quad a\not=0.$$

$$x^{2}+\frac{b}{a}\,x+\frac{c}{a}=0$$

$$x^{2}+\frac{b}{a}\,x+\frac{b^{2}}{4a^{2}}-\frac{b^{2}}{4a^{2}}+\frac{4ac}{4a^{2}}=0$$

$$\left(x+\frac{b}{2a}\right)^{2}=\frac{b^{2}-4ac}{4a^{2}}$$

$$x+\frac{b}{2a}=\pm\frac{\sqrt{b^{2}-4ac}}{2a}$$

$$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$

**Do # 2: Only skip steps you can do in your sleep:**

$$x=\frac{2}{y}\implies y=\frac{2}{x}.$$

You can swap the LHS and the denominator of the RHS in this fashion, provided you don't then divide by zero (remember the domain issue I mentioned above!)

$$2-x=y\implies 2-y=x.$$

Similar to above.

**Do # 3: Rationalize a square root:**$$\frac{\sqrt{1+h}-\sqrt{1}}{h}=\frac{\sqrt{1+h}-\sqrt{1}}{h}\cdot\frac{\sqrt{1+h}+\sqrt{1}}{\sqrt{1+h}+\sqrt{1}}=\frac{1+h-1}{h(\sqrt{1+h}+\sqrt{1})}=\frac{1}{\sqrt{1+h}+ \sqrt{1}},$$

for $h\not=0$.

This is useful in calculus when evaluating limits.

These are all that come to mind at the moment. If you are aware of any other common mistakes or valid and useful techniques you see, please reply to this thread. I will moderate this thread, but you can help make this thread more valuable with your experience. Thank you!

Questions and comments should be posted here:

http://www.mathhelpboards.com/f49/commentary-algebra-dos-donts-4224/#post19141

Last edited by a moderator: