This has been bugging me for some time now, so I ask you to try to help me realize what is going on here. I just can't get my brain around this. I have a proper fraction and a negative integer. The fraction is added to the integer.


This admittedly looks like a very simple problem, something very trivial. My first reaction when I see a problem like this is to rewrite the integer as a fraction and then multiply the numerator and the denominator so that I have the same denominator in both fractions. Adding fractions requires the denominators to be the same.


This is a legitimate statement, and the equality still holds. But I cannot reduce the improper fraction -199/2 any more, it is already in its simplest form. But I should be able to write this as a mixed number.


This is where my brain work stops. How do I get this mixed number? Does it matter that I can't reduce the fraction -199/2 any more? I think this is what might be bugging me all this time, since it won't go into it evenly and I get a remainder. I am not used to that when writing results as mixed numbers.

I can rewrite this mixed number like this.


As you can see, the value of it, expressed in decimal form, is still the same. But I just don't see the connection between $-100+1/2$ and the mixed number $-99\frac{1}{2}$.

So how do you deal with this? Please lead me through it step by step.


I thank you all for the help you have given me. It is very much appreciated! I can finally rest my mind knowing that I have cracked this nut.

Just to let you know, the original problem was dealing with inequalities. "Check that the given number is a solution to the corresponding inequality." Problem number 4 c was to check that $y=40$ satisfies the inequality $-\frac{5}{2}y + \frac{1}{2} < -18$.

$$-\frac{5}{2}y + \frac{1}{2} < -18 | y=40$$ $$-\frac{5}{2}40 + \frac{1}{2} < -18$$ $$-\frac{5}{1}20 + \frac{1}{2} < -18$$ $$-100 + \frac{1}{2} < -18$$ $$-99 \frac{1}{2} < -18$$

This problem was presented by Sal over at Khan Academy. He is usually very thorough when explaining how to solve some equation, or inequality in this case. He will usually explain every step. But in this video he was trying to squeeze in as many problems as possible in the least amount of time. So I guess he skipped a step or two. But I didn't understand how he got that mixed number on the left side. Now I do! Thank you!

Now, the way I understand it, there are two ways to get from $-100+\frac{1}{2}$ to $-99\frac{1}{2}$. One way is to get the improper fraction and do the long division. Another way, or the "trick" that I think Sal uses here, is to split the numerator into two numbers.

Long division

$$-100+\frac{1}{2}=\frac{-100}{1}+\frac{1}{2}=\frac{-100 \cdot 2}{1 \cdot 2}+\frac{1}{2}=\frac{-200}{2}+\frac{1}{2}=\frac{-200+1}{2}=\frac{-199}{2}=-99\frac{1}{2}$$

When you arrive at $\frac{-199}{2}$ you need to do the long division. You divide 2 into 199. And you get that 2 goes into 199 evenly 99 times with the remainder 1. So 99 is the whole number part and the fraction is 1/2 where remainder 1 is the numerator and 2 is the denominator. The whole thing is of course negative so you put a negative sign in front of it.

Splitting the numerator


So the trick here is to rewrite or to "split" -199 in the numerator into two different numbers, so that one of them becomes even (divisible by 2) and when you sum them up you get -199.

You can add -1 to -198 and it becomes -199. Adding a negative is the same thing as subtracting the positive of that number, so this really becomes -198-1 which is indeed -199. So you are not changing the value of the numerator, so this is safe to do.


What this allows you to do is to write the two numbers as two separate fractions.


Since one of them is an improper fraction and is divisible by the denominator you have, which is 2 in this case, you can simplify the expression by dividing the improper fraction, and since you made sure that one of the numbers is divisible by 2 it will go into it evenly. And you are left with a whole number and a proper fraction.


There are also other numbers you could add together and get the sum -199. But remember, one of them must be even so you can divide it by 2. You could add -3 to -196. Adding -3 to -196 is the same thing as subtracting 3 from -196 which is also -199.


So it's the same thing here.


Sure, -196 is divisible by 2, but the problem here is that you also have the improper fraction $\frac{-3}{2}$.


So what this tells you is that you can divide 2 into -196 one more time. So notice that $-99\frac{1}{2}$ is the same thing as $-98\frac{3}{2}$. And if you keep on doing this you will see that this is the same thing $-97\frac{5}{2}$ and this in turn is the same thing as $-96\frac{7}{2}$ and so on.


As you can see a pattern is emerging here. But the simplest of these numbers is the $-99\frac{1}{2}$ and this is the preferred answer for a mixed number.

Wow! That was much more than I wanted to write. I just thought I would share my findings, but I got carried away. The bottom line is that you can handle -100+1/2 like a regular fractions problem. And when you arrive at -199/2 there are two paths you can take. You can either do the long division and write the remainder as the fractional part of the mixed number, or you can rewrite the non-even numerator as a sum of one even number and one non-even number (the remainder). That was the trick I guess, that Sal pulled on me.

It's worth noting how the numerator of the fractional part of the mixed numbers in my last example are all odd (non-even) numbers like 1, 3, 5, etc.

It's problems like these where I have to think hard that really make my math brain grow. Hopefully someone will find this "essay" helpful.

  • $\begingroup$ First of all, my statement about not being used to writing results as mixed numbers when there is a remainder is false. I must have a remainder, that's what makes the mixed number expression useful. Secondly, when doing long division of 199/2 I need to stop dividing when I get to the remainder 1. Previously I kept on adding zeros after the decimal point and got the exact result of 99.5. $\endgroup$
    – Samir
    Feb 25, 2013 at 11:31
  • $\begingroup$ Yes you stop at 1, if want to get a mixed number. The remainder 1 indicates that you add 1 /2 to 99. $\endgroup$
    – Kasper
    Feb 25, 2013 at 12:02
  • 5
    $\begingroup$ We complain a lot here about people who ask elementary questions without explaining what they tried or where they got stuck. Thank you for doing the opposite of that. $\endgroup$
    – MJD
    Feb 25, 2013 at 13:05

4 Answers 4


Please lead me through it step by step. \begin{align*} -100+\frac{1}{2} &= \frac{-200}{2}+\frac{1}{2}\\ &= \frac{-200+1}{2} \\ &=\frac{-199}{2}\\ &=-\frac{199}{2}\\ &= -\frac{198+1}{2}\\ &= -\left(\frac{198}{2}+\frac 12\right)\\ &= -\left(99+\frac{1}{2}\right)\\ &= -99\frac12 \end{align*}

  • 3
    $\begingroup$ I accept this as the answer to my question because this is the way I probably would have solved this problem. It's just that I didn't realize that I can rewrite the numerator -199 as the sum of one even and one odd integer. That was the tricky part of it. This is precisely the kind of answer I was looking for. I find it most useful to give concrete examples than to try to describe mathematical relations in words. $\endgroup$
    – Samir
    Feb 25, 2013 at 20:24

The unary $-$, i.e. the sign you put in front of numbers to show that they're negative has an explicit meaning. "$-99\frac{1}{2}$" means "The unique number that is such that if you add it to $99\frac{1}{2}$ you get $0$". This is a property shared by $-100 + \frac{1}{2}$ and thus they have to be the same number.

If you want concrete calculations, then we can do it like this: $$ -100 + \frac{1}{2} = -(100 - \frac{1}{2}) = -(99 + \frac{1}{2}) = -(99\frac{1}{2}) = -99\frac{1}{2} $$

  • $\begingroup$ So your first statement is saying that $99\frac{1}{2}+(-99+\frac{1}{2})=0$ right? $\endgroup$
    – Samir
    Feb 25, 2013 at 11:48
  • 1
    $\begingroup$ @Sammy I think it is: $99\frac{1}{2}+-(99+\frac{1}{2})=0$ $\endgroup$
    – Kasper
    Feb 25, 2013 at 12:05
  • 1
    $\begingroup$ @Kasper Yes, thank you fro pointing that out! That is what I meant. My original line of thought was $99 \frac{1}{2}+(-(99\frac {1}{2}))=0$ and this really simplifies to what you just wrote. I will simplify it right here. $$\begin{align}99\frac{1}{2}+(-(99\frac{1}{2})) &=99+\frac{1}{2}+(-(99+\frac{1}{2}))\\ &=99+\frac{1}{2}+(-99-\frac{1}{2})\\ &=99+\frac{1}{2}+(-99)-\frac{1}{2}\\ &=99+\frac{1}{2}-99-\frac{1}{2}\\ &=99-99+\frac{1}{2}-\frac{1}{2}\\ &=0+\frac{0}{2}\\ &=0+0\\ &=0\end{align}$$ $\endgroup$
    – Samir
    Feb 25, 2013 at 14:44
  • $\begingroup$ Oops! I don't know how that 2 in the denominator in step 3 and step 5 got up there. It should sit below the 1 and the division line as in $\frac{1}{2}$. I would try to use the \over tag within curly brackets, but I can't edit it now. But you get the idea. $\endgroup$
    – Samir
    Feb 25, 2013 at 15:00
  • $\begingroup$ In that second statement I think that $-(100 - \frac{1}{2}) = -(99 + \frac{1}{2})$ was hardest for me to realize. What really helped me understand this was drawing a number line and plotting those two numbers on the line. So the 99+1/2 is halfway between 99 and 100 which is why you add 1/2 to 99. All of this is negative of course so you have to distribute the negative sign and you get $-99 \frac{1}{2}$. But I only considered the positive side of it. $\endgroup$
    – Samir
    Feb 25, 2013 at 15:15

You cannot reduce the simple fraction $\,\frac{7}{2}\,$ any more, yet you can write it as the improper one $\,3\frac{1}{2}\,$...how? Just divide with residue $\,7\,$ by $\,2\,$. The quotient is, naturally, the integer part, the residue the fractional part's denominator.

The same exactly do with $\,-\frac{199}{2}\,$: quotient $\,=99\,$ , residue $\,=1\,$ , so


  • $\begingroup$ "The quotient is, naturally, the integer part, the residue the fractional part's denominator." I think you meant to say that the residue is the fractional part's numerator. And by residue you mean remainder. The remainder of $\frac{7}{2}$ is 1 and that's what you put in the numerator. The denominator stays the same as before, it is still 2. So 2 goes into 7 three times evenly, with the remainder 1. So this becomes $3\frac{1}{2}$. $\endgroup$
    – Samir
    Feb 25, 2013 at 15:48
  • $\begingroup$ These same rules apply to the fraction $-\frac{199}{2}=\frac{-199}{2}$. You are absolutely right. This is the general method for converting improper fractions to mixed numbers. I had to review this a little bit. Thanks for reminding me! $\endgroup$
    – Samir
    Feb 25, 2013 at 15:51

The question isn’t entirely clear; I’m interpreting it as asking for an intuitive justification for the direct conversion of the fraction $\frac{-199}2$ to the mixed number $-99\frac12$. You could think of it this way: $\frac{-198}2=-99$, and $\frac{-200}2=-100$, and clearly $\frac{-199}2$ is halfway between these, since $-199$ is halfway between $-198$ and $-200$. To get from $-99$ to $-100$, you must subtract one, so to get halfway from $-99$ to $-100$ you must subtract $\frac12$, getting $-99-\frac12=-99\frac12$.

Alternatively, you could use a very little algebra initially to move the minus sign:


This has the effect of letting you work with positive numbers, which apparently cause less trouble, and only shift to negative numbers at the end. If you don’t want to do that, you can still write


(I’m offering several possible ways of looking at it, because I really can’t tell quite what’s bothering you.)

  • $\begingroup$ Thank you! I think what was bothering me is how you can go directly from $-100+\frac{1}{2}$ to the mixed number $-99\frac{1}{2}$ without rewriting the -100 as a fraction first, then multiplying it to get the same denominator as the fraction 1/2 and then adding these two together, and finally getting the improper fraction -199/2. How you can just jump from $-100+\frac{1}{2}$ to $-99\frac{1}{2}$ in two seconds without ever revealing the steps you take to get there. But I think I understand now. The trick is really to split the numerator into two numbers. I'll explain in my original post. $\endgroup$
    – Samir
    Feb 25, 2013 at 16:42
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    $\begingroup$ @Sammy: Or you can think in terms of movement along the number line: in those terms $-100+\frac12$ says start at $0$, go left $100$ units, and then go right $\frac12$ unit. Since the second move is in the opposite direction from the first, it ‘backs you up’ half a unit to $-99\frac12$. $\endgroup$ Feb 25, 2013 at 16:44
  • $\begingroup$ Yes, in situations like this it really helps to plot the numbers on the number line to help you visualize what's going on. $\endgroup$
    – Samir
    Feb 25, 2013 at 16:47

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