I thought you could only mod positive numbers but then I saw this
and became confused…How does this even work ? how can you have negatives?
$$\begin{align*}
-8 &\equiv 7 \pmod{5}\\
2 &\equiv -3 \pmod{5}\\
-3 &\equiv -8\pmod{5}
\end{align*}$$
Actually….What I thought was that mod just means how many times a number goes into another number while leaving a remainder…
But after some reading I find that negative values can be equal to positive values as long as some value-remainder is divisible by the modulus. But it doesnt make sense to me that the remainderi n these equation is greater than the modulus AND that the remainder is positive while answer on left end is negative.
Best Answer
There are two notions of "mod", but I'm afraid that your description is incorrect for both.
What you describe, "how many times a number goes into another number while leaving a remainder" is the following: If you are dividing $b$ by $a$, and you write $b = qa + r$ with $0\leq r \lt |a|$, then $r$ is called the "remainder", and $q$ is called the quotient. What you describe is $q$, the quotient.
Now, to the two notions of "modulo":
Modulo operation
This is very common in Computer Science. Here, $\bmod$ is a binary operation (like $+$, $-$, $\times$).
The precise definition varies from language to language. A common one is the following: if $a$ and $b$ are integers, and $a\neq 0$, then $b\bmod a$ (pronounced exactly like it's written) is defined to be the remainder when dividing $b$ by $a$. That is, if we write $b=qa + r$ with $0\leq r\lt|a|$, then $b\bmod a$ is defined to be $r$. Therefore, $7\bmod 3 = 1$ (because $7=2\times 3 + 1$), $43\bmod 13 = 4$ (because $42 = 3\times 13 + 4$), and $1001\bmod 13 = 0$ (because $1001 = 77\times 13 + 0$). This "mod" does indeed only give you nonnegative numbers.
There are other versions of the modulo operator; some, return the "residue class of smallest absolute value" (for example, modulo $3$ will return either $-1$, $0$, or $1$, rather than $0$, $1$, or $2$; and modulo $7$ will return $-3$, $-2$, $-1$, $0$, $1$, $2$, or $3$). Others, as robjohn mentions in comments, return a remainder of the same sign as $a$, others of the same sign as $b$.
Modulo relation
This is more common in mathematics. Given an integer $n$, we define a relation called "congruent modulo $n$" as follows: we say that two integers $a$ and $b$ are "congruent modulo $n$", written $a\equiv b \pmod{n}$, if and only if $b-a$ is a multiple of $n$. Note that $a$, $b$, and $n$ can be positive, negative, or zero.
This is what you are seeing: $-8\equiv 7\pmod{5}$ is true, because $7-(-8)=15$ is a multiple of $5$. $2\equiv -3\pmod{5}$ because $-3-2 = -5$ is a multiple of $5$. $-3\equiv -8\pmod{5}$ because $-8-(-3) = -5$ is a multiple of $5$.
Note that this is a relation: you don't get a number "out of it", like you do with the modulo operation above; this is a statement about whether two numbers are related.
Connection between the two notions
The simplest connection between the two concepts is the following:
Theorem. Let $a$ and $b$ be integers, and let $n\neq 0$ be a nonzero integer. Then $$a\equiv b\pmod{n}\textit{ if and only if } a\bmod n = b\bmod n.$$
That is: $a$ and $b$ are congruent modulo $n$ if and only if they leave the same remainder when divided by $n$.
Proof. Suppose that $a\equiv b\pmod{n}$, and write $a=qn + r$, $b=pn+s$, with $0\leq r\lt |n|$, $0\leq s\lt |n|$. Then $a\equiv b\pmod{n}$ if and only if $(qn+r)-(pn+s) = (q-p)n + (r-s)$ is a multiple of $n$; this occurs if and only if $r-s$ is a multiple of $n$, which occurs if and only if $|r-s|$ is a multiple of $n$.
But $0\leq |r-s|\lt|n|$; the only multiple of $n$ that has absolute value smaller than $|n|$ is $0$. So $r-s$ is a multiple of $n$ if and only if $|r-s|=0$, if and only if $r=s$. Since $r=a\bmod n$ and $s=b\bmod n$, we conclude that $a\equiv b\pmod{n}$ if and only if $a\bmod n = b\bmod n$. $\Box$