# Incredible frequency of careless mistakes [closed]

Ever since high school, I've had a serious problem with math classes. Be it discrete math, algebra, calculus or linear algebra, I seldom have trouble understanding the texts or lectures, but when it comes to executing calculations on homework and exams, I can never do it: I keep making mistakes. It's been a constant source of frustration, and right now, I'm on the verge of failing linear algebra and second year calculus (which I've been avoiding for as long as I could).

I'm currently going through all the exercises and videos on Khan Academy, beginning with one-digit addition, working my way back up to integrals. I was making a lot of mistakes with multiple digit subtraction and multiplication, so initially, I was hopeful that maybe with enough practice in those two things, I'd be good to go again. But it turns out that my mistakes aren't confined to any one task, and the number of mistakes I make per exercise just accumulates as the tasks become more and more computationally involved:

• Thinking that 7 + 5 = 13
• Neglecting negative signs when transcribing
• Carrying in multiple-digit addition when a digit hasn't surpassed 10
• Multiplying instead of adding
• Using the wrong exponent laws, log laws, limit laws etc.
• Mixing up trigonometric function properties
• Forgetting to reverse signs when expanding negated expressions of multiple terms

A typical page from my notebook: I've tried everything: going for walks, ensuring that I have good sleep, writing out every step with almost hilarious verbosity, changing rooms, starting with fresh notebooks, reviewing everything I write, trying different notations, heck I even spent a month refining my handwriting and posture for good measure.

I'm certain I spend about twice as much time as other people practicing too. I avoid taking other courses in the same semesters that I take math courses.

As far as personality goes, in general, I'm not exactly OCD, but I'm not careless either. I probably have more patience than most people -- a bit too much, as my friends would tell me. I'm sure you can also tell from my adequate grammar, punctuation and spelling that I probably don't have any kind of learning disability that would prevent me from following symbolic rules. I've gotten near 100% in a first order logic course too.

Yet when I do math, I always screw up. It's been this way for over a decade, maybe more.

I don't know what I'm doing wrong anymore, and I'm losing hope.

What could I do?

## closed as off-topic by user223391, Taroccoesbrocco, Mostafa Ayaz, Chris Custer, Rafa BudríaJul 14 '18 at 20:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – Community, Taroccoesbrocco, Mostafa Ayaz, Chris Custer, Rafa Budría
If this question can be reworded to fit the rules in the help center, please edit the question.

• Have you gotten a full physical, including a neurological workup? – marty cohen Sep 28 '11 at 18:45
• Dyscalculia sounds like a possibility; it is not incompatible with excellent writing skills. It might be worth your while to get an evaluation from an expert in learning disorders. – Brian M. Scott Sep 28 '11 at 18:54
• @Brian Some of those symptoms are familiar, though probably not to as severe a degree as Wiki seems to suggest it should be. If I do have dyscalculia, it's probably not a result of working memory problems: I scored off the charts in numerical memory when I had my IQ tested in elementary and middle school. Though I'm certain my IQ's dropped quite a bit since then, I think arithmetic was frustrating for me even back then. – Rei Miyasaka Sep 28 '11 at 19:48
• I’d guess that if dyscalculia is the problem, it’s in a relatively mild form, but I really know very little about it; I’ve just had students who had been diagnosed with it. – Brian M. Scott Sep 28 '11 at 19:59
• Should this be CW? – JavaMan Sep 28 '11 at 21:06

## 7 Answers

@Rei, I've gone through pretty much the same problems as you in school, and also made the same sort of mistakes! And I'll assume yours isn't a medical problem. You mentioned a big list of mistakes as being your problem. You probably need to ask yourself first: is there something fundamentally wrong in your understanding, or is it that you just make mistakes when pressed for time, etc.? If it's the former, get the concept right first, saving the calculations for later. If it's the latter, it's a much more common problem that we all get into. I can't give you a more specific answer in that case without knowing more about when you make these careless mistakes.

If that seems too abstract, here's some concrete help about your $x=480\times 72$ problem. Don't do it multiple times in the same way - instead check whether the answer makes sense in different ways! (This is an extremely important step that I've seen many people neglect, because they get a correct answer. That's really not understanding maths.) First off , $x = (480 \times 72) > (y=480 \times 70)$, where $y$ should be very easy to do :$y= 48 \times 7 \times 100 = 33600$. This alone should tell you that , if you got the correct answer in your worksheet, it must be 34560!

Next step: (I assume you know basic algebra) We know that $x = 480 \times 70 + 480 \times 2 = 480 \times 70 + 960$. If that seems too difficult to compute, just take $x \approx 480 \times 70 + 1000 = 34600$. Now, we finally get your answer of $34560$, by subtracting 40!

Ok, suppose you're still not sure. That's fine, we'll do another quick check. It's easier $z= 500 \times 70$ which is just $5\times 7$ with three zeros to follow, so $z=35000$. And $x$ should be a little less; if you're interested you can find the difference. Otherwise, you can check your computations by finding which answer you got was closest to $35000$, and discarding the rest.

What have we gained in the process?

• We admitted that certain computations are hard to do by hand.
• Hence, we looked for concepts to simplify those computations, and looked for approximate answers.
• We checked our answers several times, in multiple ways.

Here're some thumb rules to keep in mind:

• Human beings make errors in a variety of situations, so you're no exception!
• Since you mentioned errors in multi-digit addition, subtraction , etc. you may want to try those first,ignoring everything beyond.It's not important that you compute very fast, but it's very important that you get the algorithms very clearly.
• When you get an answer, just stop and ask, "how can I check this"? Check your answer any way you want - use a calculator, computer, etc.

Here're some simple exercises to try out; you may want to try finding an approximate answer first, without doing any pen-and-paper calculation:

• $99 \times 99$
• $3.14 \times 2.99$
• $-1 - (-1)$

As an example, I'll give some hints for the first one. Let $a=99^2$. I find it hard to square 99, so I'll just square 100 and say $a$ is slightly less than 10000. Next, I'll maybe use an identity like $a=(100-1)^2$ and zoom in on how much $a$ is less than 10000. I'll perform a crude check by stating that $a>90 \times 90 = 8100$. So $a$ lies between $8100$ and $10000$.

Hope I've taken the keen edge off your despair!

Does that help?

Try working out problems with someone else who can watch over your shoulder and point out mistakes in real time as you make them. Then you can correct the mistake right away, before it propagates into rest of your work. This will allow you to notice and appreciate incremental improvements where you need fewer corrections, as opposed to having one less reason your final answer was wrong.

It may also help to reverse the process, to look for mistakes others make as they work, to train your ability to notice mistakes, which you can then apply to your own work.

• That might help. Thanks. – Rei Miyasaka Sep 29 '11 at 1:42
• "To look for mistakes others make as they work, to train your ability to notice mistakes." How would spotting mistakes make me better at spotting mistakes? What would I learn from it? – Kelmikra Oct 26 '15 at 8:12
• @Kelmikra Watching somebody else do something puts you in a different frame of mind than when you do it yourself. Also practicing spotting mistakes will help you spot mistakes because practicing is how you get better at any skill. – Joseph Garvin Aug 6 '17 at 20:15

If you know you are having certain problems, try focus on the specific problem. Also, don't work on the problem in isolation -- work on it in context too.

Also, work out your own way of doing things. From looking at your notebook paper, my first instinct is that part of your difficulty is "skipping steps" -- trying to do several steps in your head when you still have difficulty with the individual steps is a surefire recipe for disaster.

Look at your work there -- all of your errors occur in the same spot, where it looks like you are doing two things:

• You are multiplying $4 \times 7$ in your head
• You are adding $5$ to the result (because it carries over from $8 \times 7$)

You may have better luck working out the problem in a way that doesn't require you to do two things at once. For example, you might compute $480 \times 7$ separately in a separate diagram: $$\begin{matrix} & & 4 & 8 & 0 \\ \times & & & & 7 \\ \hline & & & & 0 \\ & & ? & ? & \\ + & ? & ? \\ \hline & ? & ? & ? & ? & \end{matrix}$$

Or you might try "lattice multiplication" which lets you do the individual single-digit products independently of the addition steps.

• Skipping steps is definitely a habit that I have. I try my best to shake it. Sometimes I have a better success rate when I try very hard not to skip steps; sometimes it doesn't make much of a difference. – Rei Miyasaka Sep 28 '11 at 19:56
• "trying to do several steps in your head when you still have difficulty with the individual steps is a surefire recipe for disaster." I make an incredibly large number of errors, but I haven't noticed that doing multiple steps at once increases the rate of errors. It does, however, make me much faster, so I can't call it a "surefire recipe for disaster." – Kelmikra Oct 26 '15 at 8:15
• @Kelmikra: The error rate for most people multiplying 10 digit numbers in their head will be higher than multiplying 2 digit numbers in their head. More steps means holding more in your working memory for longer. It is guaranteed to be more error prone in the limit. – Joseph Garvin Aug 6 '17 at 20:17

Six years later, I have an answer. This is going to ruffle some feathers, but:

I went vegan. I kid you not.

Mostly, going vegan was to try to see if it would improve my skin (which it did), but this seems to have been another positive effect. Aside from the acne, I've been in pretty good health and shape for my whole life -- good BMI at 20.5, good cholesterol levels, blood pressure etc. -- so I never really suspected anything.

I've been testing myself on-and-off every few months or so on Khan Academy, because I'd often read computer science papers and find myself wishing I could learn math properly. A few months since I changed my diet, I tried some particularly arithmetic-heavy exercises on Khan (systems of equations, long multiplication and division, matrix multiplication etc.), and I've been nailing everything.

It could be that I'm mildly allergic to animal products, or it could be the result of a change in gut flora having an effect on my nervous system. Or it could be the oil or sodium or iron or contaminants that I happened to remove from my diet along with animal products. Every now and then, but less often recently, I still have whatever looks good on the menu when I'm out with friends, but not without resigning to the fact that my face will look like Minecraft by the next day.

I've ruled out most other environmental issues as well as things like stress and sleep.

Without getting into any of the politics or ethics or health debate regarding veganism, I would recommend trying some kind of elimination diet to anyone who's having inexplicable problems.

Personally I spent a couple of weeks improving my fundamental math skills on Khan Academy.

I practiced mentally doing 1-digit addition, then 1-digit subtraction, then multiplication, then continued with 2-digit calculations.

It's like any other skill. If your fundament is bad you will not be able to excell. Define your challenges - on paper - and work through the most fundamental problems until it sits real tight. Hehe..

And don't do too many steps in your head at once (unless your fundament is beyond SOLID)

Come to think about it I have a neat little theory. It is that we are not limited in the amount of information we can process at a given moment, but that it deterioates fast. Think about how much information you process subconsciously at any given moment.

Combine that with the fact that we myelinate our brains when we practice something, which in turn increases the speed and precision at which the neurons are able to fire signals and you have a REALLY strong argument for improving the fundamental mathematical skills and practicing them regularly!

It depends on what you mean by a mistake. If you define mistake as something that will lose you marks on a math test and you already know how to show your work and you sometimes slip up and make a mistake that you later can see is obvious why it's a mistake, maybe you could get a friend to watch you like a hawk while you're doing your homework and notify you immediately when you make a careless mistake. Then you should take it seriously making the effort to make those careless mistakes less often. According to https://www.youtube.com/watch?v=_St8gcqPB7E, one error almost everybody makes when they're adding mentally is thinking that 1000 + 20 + 30 + 1000 + 1030 + 1000 + 20 = 5000 when it really equals 4100.

That's not the only reason for making mistakes in math. According to https://www.youtube.com/watch?v=1irvvZzbJkU, a lot of people can't find the mistake in proving 2 = 0 as follows: $2 = 1 + 1 = 1 + \sqrt1 = 1 + \sqrt{(-1)^2} = 1 + \sqrt{-1}\sqrt{-1} = 1 + i^2 = 1 + (-1) = 0$ The mistake can be avoided by going down to the basics. For example, you could decide how you want to formalize a certain statement in Zermelo-Fraenkel set theory. According to this answer, instead of just declaring that certain properties of the real numbers are true and proving theorems from them, you can instead construct a set with operations in Zermelo-Fraenkel set theory and define it to be the set of all real numbers and prove that that set with those operations is a complete ordered field. It probably turns out that the standard formalization of the square root function on a complex number in $\mathbb C$ is to pick the complex number whose real part is positive and whose square is that number or if the original number was a zero or negative real number, to pick the number whose square is that number and has a nonnegative imaginary part. I beleive that complex exponentation can be defined as follows: $\forall x \in \mathbb{C} \forall y \in \mathbb{C}$

• $0^x = 0 \text{ if } re(x) > 0$
• $0^x = 1 \text{ if } x = 0$
• $0^x \text{is undefined otherwise}$
• $\text{ if } x \in \mathbb{R} \text{ and } y \in \mathbb{Q} \text{ and }y\text{ has an odd denominator in lowest terms and } x < 0, \text{ then } x^y = exp(ln(-x) \times y) \text{ if the numerator is even and } -(exp(ln(-x) \times y)) \text{ if the numerator is odd }$
• $\text{ otherwise } x^y = exp(ln(x) \times y)$

where $\forall x \in \mathbb{C}$, ln(x) is defined to be the complex number $y$ such that $exp(y) = x$ and $\frac{-\pi}{2} < \text{im}(y) \leq \frac{\pi}{2}$

Chances are in a Calculus course, you will be expected to talk about only real numbers. If a Calculus question asks you to find the set of all possible values of x for an equation with one variable x, you need to find a method that always works. If your teacher taught the class their own method that always works and you use your own method and get the right answer, you might still not get full marks because you didn't show your work using their method. You should recognize what the inverses of various functions mean. The exp function is injective, that is, it assigns a different number to every number so when ever a number is in the range of exp, its inverse of exp, ln of that number, is the one number that exp assigns that number to so if ln x = y, exp y = x. Other functions like the squaring function, sin, and cos are not injective so for each number in the range of that function, its inverse applied to that number is just one of the numbers that that function assigns that number to. When a number is in the range of the squaring function, its square root is the nonnegative number whose square is that number so if y = sqrt(x), x = y^2. For any number between 1 and -1, sin^-1 of that number is the number between π/2 and -π/2 such that the sin of it is that number so if y = sin^-1(x), x = sin(y). For any number between 1 and -1, cos^-1 of that number is the number between 0 and π such that the cos of it is that number so if y = cos^-1(x), x = cos (y). One possible method a teacher could have is having a set of allowable operations on a one variable equation, some of which never gain or lose any solutions and some of which sometimes gain solutions but never lose solutions; requiring that each equation is the result of the application of one of those operations to the equation on the previous line and that unless every single one of those operations is one that never gains or loses any solutions, you have to check each solution to the final equation to see if it is a solution to the original equation in order to get full marks on that question. Suppose f, g, and h are expressions involving the only the variable x. Here are some operations that they and their inverses never gain or lose any solutions when using real numbers. For any x, the expression f ≠ g is true if and only if x is a number such that f and g are both defined and are not equal to each other. Adding the negation symbol ¬ to the beginning of an expression changes the expression to an expression that means the original expression is not true.

f = g to g = f

f^3 = g^3 to f = g

f = g to e^f = e^g

f = g to ln e^f = g

f = e^g to g = ln f

f^2 = g^2 to (f = g or f = -g)

f ÷ g = h to (h × g = f & g ≠ 0)

f + g = h to e^f × e^g = e^h

$f = sin(g) \text{ and } \frac{-\pi}{2} < g < \frac{\pi}{2}$ to $g = sin^{-1}(f)$

$f = cos(g) \text{ and } 0 < g < \pi$ to $g = cos^{-1}(f)$

any conversion of a term of one the following forms or its inverse

• a + b to b + a
• (a + b) + c to a + (b + c)
• a × b to b × a
• (a × b) × c to a × (b × c)
• a × (b + c) to (a × b) + (a × c)
• a^b × a^c to a^(b + c) for any positive a
• (a^b)^c to a^(b × c) for any positive a

Here are some operations that sometimes gain solutions but never lose solutions.

f = g to f^2 = g^2

f ÷ g = h to h × g = f

f × g = h to ¬(ln f + ln g ≠ ln h)

ln f = ln g to f = g

f = g to ¬(f ≠ g)

¬(f ≠ g) to ¬(ln f ≠ ln g)

$f = sin^{-1}(g)$ to g = sin(f)

$f = cos^{-1}(g)$ to g = cos(f)

any conversion of a term of one of the following forms

• ln a + log b to ln (a × b)

• ln (a × b) to ln a + ln b for positive a and b

• (ln a) × b to ln (a^b)

• ln (a^b) to (ln a) × b for positive a

Maybe in a stronger subtheory of ZF than the subtheory that lets you prove anything you're supposed to be able to prove in Calculus for full marks, given a system of equations, you can do something like deduce from the statements $\forall x \in \mathbb{R}, \text{ if }f(x) \neq 0 \text{ then } g(x) \neq 0$ and $\forall x \in \mathbb{R}, \text{ if }f(x) = 0 \text{ then } g(x) \neq 0$ and $\forall x \in \mathbb{R} f(x) \times g(x) = h(x)$, the statement $\forall x \in {R} g(x) \neq 0$ and then from that $\forall x \in \mathbb{R} h(x) ÷ g(x) = f(x)$

Here are some other mistakes people could make:

• Deducing from the statement that a real natural number is not a finite Von-Neumann ordinal and the statement that the cardinality of a finite set is a natural number the statement that the cardinality of a finite set is not a finite Von-Neumann ordinal.
• Assuming that the cardinality of any set is its initial ordinal in ZF when cardinality is only defined that way in ZFC like in this question.
• Deducing from the statements that the cardinality of any well-orderable set is its initial ordinal and the cardinality of any set is Scott's definition that the cardinality of the empty set is the Von-Neumann ordinals 0 and 1 as described in this answer.
• Can somebody please black out the numbers 5000 and 4100 for me and teach me how then I will delete this comment? – Timothy Jun 30 '17 at 18:14
• @T Enclose them in double *'s or _'s: viz. **5000**, __4100__. – Antonio DJC Feb 16 at 7:54