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So I only have an Algebra II level understanding of math seeing as I am still in high school and am still missing some fundamentals seeing as I didn't pay attention in math until this year. However when recalling something my algebra teacher had taught me during the year I came up with some questions regarding the logic recently.

So during the school year, I was taught that $\frac{2}2=1, \frac{a}a=1, \frac{xy}{xy}=1$ and so forth but $\frac{0}0= \text{Undefined}$... and while researching this topic I found that the algebraic way to write all these fractions is as such $2(x)=2, a(x)=a,$ and $0(x)=0$ and upon researching this further I found that the reason that $\frac00$ is undefined is that for any value of $x$ the equation holds true. However, seeing as in the fraction $\frac{a}a$ $a$ is a variable and variables can represent any given quantity I was wondering in the case that $a=0$ would $\frac{a}a$ still $=1$ and if not why along with the fact that lets say $a=0$ and you didn't know it why is it safe to assume that $a$ would never equal zero? Also if it happens to be the case where when $a=0, \frac{a}a=1$ (which I doubt it is) shouldn't this mean that $\frac{0}0=1$ then?

Siong Thye Goh
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Matthew B
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    Anytime variables appear in denominators we automatically exclude the case in which it is zero because it doesn't make sense. – Sean Roberson Jul 26 '17 at 17:40
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    @SeanRoberson I'd be careful with that generalization. There are many applications where finding zeros in the denominator is important, the first I remember learning being asymptotes and holes, which shouldn't be too far from where OP is. Sure, usually we don't care about the zero case, but it's important to look at the context of the question before assuming we can ignore it. In response to why is it safe to assume that a would never equal zero, I'd say technically it's not. But mathematicians are notoriously lazy, so if it's trivial to say "unless a is 0", they'll probably just ignore it. – Lord Farquaad Jul 26 '17 at 20:47
  • There are subtleties about precisely what an expression like $a/a$ actually means. The subtleties don't matter for most things which is why they tend not to be discussed -- but they matter a lot in this specific edge case. Alas, since they tend not to be discussed, this issue tends to be treated in an ad-hoc fashion. :( –  Jul 26 '17 at 22:54

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$$\frac{a}{a} = \begin{cases} 1 & a \neq 0 \\ \text{undefined} & a = 0\end{cases}$$

Strictly speaking, you have to check whether $a$ is zero before you cancel terms off from both numerator or denominator.

Even before we write down $\frac{a}{a}$, we should first check if the denominator can be zero, i.e. check it before we write down such term. (Credit : Sean Roberson)

Siong Thye Goh
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  • I appreciate the response, however, I have just thought of another question, let's say a= infinity would infinity/infinity = 1 or undefined also would 2(infinity) = infinity since infinity by nature encompasses all numbers? – Matthew B Jul 26 '17 at 17:52
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    $\frac{\infty}{\infty}$ would be undefined as well. You might encounter them in your calculus class in the form of indeterminate form. $2 \infty = \infty$ because multiplying something that is arbitrary huge with a positive number is still something that is arbitrarily huge. Also, a remark: $\infty$ is not a real number. – Siong Thye Goh Jul 26 '17 at 17:55
  • Once again thank you for your response and I apologize if I am being a pest but also this year in my class when describing either the domain or range that is all real numbers we wrote it in interval notation as (-infinity,infinity) thus suggesting that there is some difference between the two, as such I always assumed that -infinity encompassed all the negative numbers while infinity encompassed the positive. As such I was curious though it doesn't matter for interval notation if 0 is included in -infinity, infinity or both? – Matthew B Jul 26 '17 at 18:01
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    The notation $x \in (a,b)$ means $a < x < b$. Similarly, $x \in (-\infty, \infty)$, means $-\infty < x < \infty$. $\infty$ is not a set of numbers. We do not write $x \in \infty$. Relevant question, is $0$ positive or negative? answer is neither. but it is smaller than any positive number and bigger than any negative number. – Siong Thye Goh Jul 26 '17 at 18:11
  • Regarding $\frac{\infty}{\infty}$, is it possible that set theory has an answer via cardinal numbers, and that it depends on which infinite set one is dealing with? – Lugh Jul 26 '17 at 18:26
  • https://math.stackexchange.com/q/181304/306553 there is a discussion about $\frac{\infty}{\infty}$ here. FYI. – Siong Thye Goh Jul 26 '17 at 18:52
  • @Lugh As far as I'm aware, one never divides infinite cardinals. One might conceivably take the index of a subgroup of an infinite group, but I've never heard of dividing cardinals. – Patrick Stevens Jul 26 '17 at 21:20
  • "you have to check whether a is zero before you cancel terms off" - otherwise you get such interesting mistakes like: "Solve x² = x" --"Cancel x on both sides ==> x = 1; done!". – CompuChip Jul 26 '17 at 22:27
  • @CompuChip That is the correct solution though, isn't it? – sgf Jul 26 '17 at 22:58
  • @sgf we missed out a solution though. $x=0$ or $x=1$. – Siong Thye Goh Jul 26 '17 at 22:59
  • @SiongThyeGoh Right, I forgot about that. But the really interesting mistakes are those that produce wrong equations (like in Chickenmancer's answer), not correct solutions in cases that have more than one solution. – sgf Jul 26 '17 at 23:00
  • @sgf exactly my point! You can only "cancel x on both sides" if $x \neq 0$. You need to separately check what happens if $x$ is zero because $\frac{x^2}{x} = \frac{x}{x}$ is not equivalent to the original equation and when looking at the problematic point x = 0, you will in fact find that $x = 0$ is also a solution. Of course, the way around this is rewrite as $x^2 - x = 0$ and then instead of dividing out the $x$, factor as $x(x - 1) = 0$. – CompuChip Jul 27 '17 at 06:51
  • Also, since you referenced Chickenmancer's answer, note that this is arguably more dangerous since $1 = 2$ is obviously wrong while here you are actually silently and subtly dismissing half of your solutions. – CompuChip Jul 27 '17 at 06:55
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If you are working problems or doing proofs you do need to know that $a \ne 0$, if $a/a$ appears. Even though $a$ is a variable, it still represents something, and you have to understand what it represents well enough to know it won't be zero. Alternatively you can just say something like: for $a \ne 0$ then $a/a=1$.

Example 1. Problem: Let $a$ be number of pounds of apples, and $P$ be the price in dollars of $a$ pounds of apples plus 10 cents for a sack. What is cost per pound of apple? Answer: the cost per pound of apples (in dollars) is $(P-0.1)/a$ if $a$ is not zero. In this example it may be implicitly or explicitly known that $a \ne 0$ (eg "George buys $a$ pounds of apples" implies $a \ne 0$), in which case the qualification isn't necessary, but still doesn't hurt.

Example 2. Problem: Given that $a+b=15$, give an expression that is the ratio of $b$ to $a$ in terms of $a$. The simple answer is $ratio = \frac{15-a}{a}$. However that answer must be qualified as "for $a \ne 0$ then $ratio = \frac{15-a}{a}$", or it might even be better to say "for $0 <a \le 15$ then $ratio = \frac{15-a}{a}$", depending on the context of the problem. In this case you don't know what $a$ is, and the way I stated the problem you don't know what values $a$ can have, so you're forced to either get clarification about the problem or state your assumptions.

Χpẘ
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To address your question using algebra, you formally define $$\frac{a}{b}=(a,b)$$ where $b\ne 0.$ The reason for this last condition is how you define fractions as being equivalent. Namely, if $(c,d)$ with $d\ne 0$ you say $(a,b)=(c,d)$ when $ad=bc.$

You may then use relation to define addition and multiplication of these ordered pairs

$$(a,b)+(c,d)=(ad+bc,bd)\quad \quad (a,b)\cdot(c,d)=(ac,bd).$$

The previous should look very familiar, because it is fraction addition and multiplication.

Finally, to consider $\frac{0}{0},$ that is $(0,0),$ you find yourself in a bad situation, because you're in direct conflict with the definition of a fraction. Next thing you may try is to assume that if $(0,0)$ exists, then $(0,0)=(a,b)$ precisely when $a\cdot 0= b\cdot 0,$ so you see that any $(a,b)$ is equivalent to $(0,0),$ that is, $\frac{0}{0}$ isn't well defined. If you and I do computations in separate rooms, using the exact same rules for arithmetic, then we can both end up with different results, even if what we did was "allowed." Let's try. You do a computation with $\frac{0}{0},$ and I'll do one below, check your work against mine.

Take $$\frac{0}{0}\cdot \frac{1}{0}$$ and $$\frac{0}{0}\cdot \frac{2}{0}$$ these, according to traditional arithmetic, produce $$\frac{0\cdot 1}{0\cdot 0}\text{ and } \frac{0\cdot 2}{0\cdot 0}$$ Since $0\cdot0=0,$ we have $$\frac{0\cdot 1}{ 0}\text{ and } \frac{0\cdot 2}{ 0}$$ producing $$1\cdot \frac{0}{ 0}\text{ and } 2\cdot \frac{0}{ 0},$$ and if $\frac{0}{0}=1,$ (which is a choice!) then we arrive at the contradictory statement, $$1=2.$$ Which indicates, much lie the above statement that $(0,0)=(a,b),$ that any fraction $(a,b)$ where $b=0,$ is equivalent to any number you please.

Thus, we conclude that dividing by zero isn't well defined, and cannot be permitted with standard arithmetic.

Chickenmancer
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Your logic is very good! Another way to phrase the conclusion: it's NOT true that "$\frac{a}{a}=1$ always". The true statement is that

$\displaystyle\frac{a}{a}=1$ always (i.e. for any number $a$) when it's well-defined.

This more careful phrasing reminds us that first we have to check whether the expression is well-defined; and with a quick look we would realize that this property is not applicable to $a=0$.

zipirovich
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  • Thank you for your response, I presume that my teacher taught it this way because the vast majority of my class wouldn't ponder weather a/a=1 always or not. – Matthew B Jul 26 '17 at 18:02
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You may, at some point, learn about limits in mathematics. Limits are about looking at how particular things behave when you approach certain values - for example, the limit of the sequence $2.1, 2.01, 2.001, 2.0001, \ldots$ is just 2, whereas the limit of the sequence $2, 3, 4, 5, 6, \ldots$ does not exist (and we sometimes say the limit "is" infinity, although that's just notation and doesn't mean you should treat the infinity as a real number).

You can also have limits that involve approaching a value from two directions. For example, if we have the function $xy$ (i.e. $x$ multiplied by $y$), then as we approach the point $x = 2, y = 3$ from any possible direction then we always approach the value 6. It doesn't matter if we just set $x = 2$ and then try $y = 2.9, 2.99, 2.999, 2.9999, \ldots$, or if we come via the pairs $(1.9, 3.1), (1.99, 3.01), (1.999, 3.001), \ldots$ or whatever, it always gets closer and closer to being 6.

However, if we have the function $\frac{x}{y}$ and want to find the limit at the point (0, 0), then it matters very much what direction we approach from. In fact, if there's a value you want to get as the limit, there's a direction you can approach from where the limit will appear to be that value - come along the $x = 0$ line, and the limit is just 0. But try coming from pairs like $(10, 10), (1, 1), (0.1, 0.1), (0.01, 0.01), \ldots$ and the limit is 1. Come from another direction, and the limit is -10, or a googol. There's even a way to take the limit so that it becomes infinite. Since we can assign any possible limit value, we can't pick one meaningful value - and so we say that the function $\frac{x}{y}$ has indeterminate value at the point $(0, 0)$, because we can't meaningfully determine what value it should have.

ConMan
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