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Throughout school we are taught that when something is multiplied by 1, it equals itself.

But as I am learning about higher level mathematics, I am understanding that not everything is as black and white as that (like how infinity multiplied by zero isn't as simple as it seems).

Is there anything in higher-level, more advanced, mathematics that when multiplied by one does not equal itself?

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If the Multiplicative Identity is something else then 1 then there you have it. – saturatedexpo yesterday
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One thing that is sort of an example is when you manipulate some complicated expression by multiplying by $\frac{f(x)}{f(x)}$. This does equal $1$ everywhere where $f(x)\ne 0$, but it is of course undefined where $f(x)=0$. So sometimes we change the domain of our function by multiplying by this version of "$1$". – Bye_World yesterday
    
Somewhat pedantically, something is always equal to itself, even if it were say multiplied by $7$. What is different in that case is the result of the multiplication, not the something being multiplied. Which is why I would say "when something is multiplied by $1$, the result equals the original thing". But of course the looser formulation is common and poses no difficulty. – Marc van Leeuwen 19 hours ago
up vote 18 down vote accepted

Usually, if there is an operation called multiplication, it is defined as having an identity element. We call that element $1$. When we do that, we define $1\times x = x \times 1 = x$. Sometimes we only define one of the equalities because we have the power to derive the other one. If we don't have a $1$, we don't have a multiplicative identity.

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So it is intrinsic to the operation of multiplication? – Anonymous yesterday
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@Anonymous Pretty much. The $1$ is the identity, which might not be $1$. Look up what a mathematical group is. – Carl Schildkraut yesterday
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You have to look at the definition of the operation. As a kid, the only multiplication is on the naturals and $1$ is the identity. Later we extend that to the rationals, the reals, and the complex numbers. In all of those cases, $1$ is the identity. We can also define arbitrary operations on arbitrary sets. When we get to that stage, it would be perverse to call an operation multiplication that did not have an identity and you would call that identity $1$. – Ross Millikan yesterday
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"So it is intrinsic to the operation of multiplication?" - I'd rather say it's intrinsic to the 1: The 1 is usually defined exactly as the element that has the property of not changing other elements when being multiplied with them (and I don't know this as "identity element", but rather as the neutral element - similar to 0 being the neutral element for addition) – Marco13 22 hours ago
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@OFRBG: And even with matrices, the identity in a matrix ring of interest is often called $1$. e.g. on the bottom of page 2 of Matrix Rings in my copy of Jacobson's algebra 1, quote, We define the unit matrix $1$ by $$1 = \left( \begin{matrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & 1 \end{matrix} \right)$$ endquote. It may look odd at first, but it really doesn't lead to confusion, since nearly any situation where you might interpret $r$ either as a scalar or as a scalar matrix, expressions involving $r$ mean the same thing either way. – Hurkyl 10 hours ago

As Ross Millikan has pointed out, mathematicians like to call multiplicative identities $1$ because we like things to behave familiarly.

There are exceptions to this though, especially when we're working with sets of numbers that are endowed with structures other than the ones we usually use. The easiest example that comes to mind is the set $S=\{0,2,4,6,8\}$ with addition and multiplication defined modulo $10$ (in other words, if I multiply or add numbers and they exceed $10$, then I take the remainder dividing by $10$. So $6+8\equiv 4$ modulo $10$ since $14=10+4$, while $4\cdot 8\equiv 2$ modulo $10$ since $4\cdot 8=32=3\cdot 10+2.$) If you sit down and multiply each element of $S$ by $6$, you will find that $6$ is the multiplicative identity, not $1$ (which isn't even in the set).

This is kind of cheating though because in some sense $S$ with this structure is "the same" as $\{0,1,2,3,4\}$ with addition and multiplication defined modulo $5$, and in this case $1$ is is the multiplicative identity.

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This is a very good point. The field axioms say there must be a distinguished element $1$ which is the multiplicative identity. You have a field and the distinguished element is named $6$. I could also define a five element field with elements $\{0, george, bill, jim, steve\}$ As long as I get the addition and multiplication tables right it will work. As you indicate in the last paragraph, there will be an isomorphism onto the field of integers modulo $5$. Depending on how I match them up, any of the boys could be $1$, the multiplicative identity. – Ross Millikan 23 hours ago
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While interesting, this does not answer the original question. That question was whether multiplication by $1$ ever does not give back the original value, not whether there are other things than $1$ that have the property that multiplication by them is the identity. – Marc van Leeuwen 19 hours ago

Outside of the usual integers / rational numbers / real numbers that we learn about in grade school, the symbol "$1$" is often used as a placeholder for the multiplicative identity in a given algebraic structure, such as a ring or field.

To build such a structure, we start with some set, which we'll call $S$. Next, we define "operations" on this set, which are functions that input any two elements of $S$ and output a single element in $S$. Typically, we call these operations "addition" and "multiplication" and use the standard $\times$ and $+$ symbols (though these can be defined quite differently from grade school addition/multiplication). Finally, the mathematician will specify a list of axioms that this set and the operations should satisfy, such as associativity, commutativity, and so forth.

Among these axioms will often be a statement along the lines of "There exists a special element in $S$, which we will call the multiplicative identity and denote by the symbol "$1$" which has the property that $y \times 1 = 1 \times y = y$ for any element $y \in S$.

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I don't think there's really any context where we want to relax the axiom $1a = a$. But sometimes, we want to relax the axiom $0a=0$. I've seen the terminology almost-semiring used to describe algebraic structures where addition and multiplication behaves as expected, except that $0a = 0$ might not be true. Almost-semirings arise naturally: for example, suppose $X$ is a set, $S$ is a semiring, and consider the collection of all partial functions $X \rightarrow S$. This is an almost-semiring, but if $X$ is non-empty, it won't satisfy $0a=0$. To see this, let $a$ denote any non-total function $X \rightarrow S$. The big difference between $1a=a$ and $0a=0$ is that $a$ occurs exactly once on each side of $1a=a$ (it's a "balanced" identity), whereas this isn't true of $0a=0$. This means that the identity $1a=a$ can be expressed using operads, whereas $0a=0$ cannot. The end result is that while most ways of building constructions will preserve the identity $1a=a$, many will not preserve $0a=0$. For another, similar example, try doing algebra in the powerset of a semiring. You'll quickly notice that $1A=A$ holds, but that $0A=0$ doesn't.

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It may not be what you're looking for, but I believe this is related. In the case of limits, repeated multiplication by a limiting value of $1$ can have surprising effects.

Consider this: $$\left(1+\frac{1}{n}\right)^{n}=\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\ldots$$

For large values of $n$, this appears to approach: $$\left(1+0\right)\left(1+0\right)\left(1+0\right)\ldots=1\cdot1\cdot1\ldots=1$$ $$1^{\infty}\stackrel{?}{=}1$$

However, this is not the true value of the limit. In reality: $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e\gt1$$

It appears that infinite repeated multiplication by $1$ somehow becomes another value entirely. This is why $1^\infty$ is considered an indeterminate form.

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It's repeated multiplication by a value approaching 1. Not by 1. – immibis 7 hours ago
    
@immibis Yes, that's why I said it appears that way. – Curtis Bechtel 7 hours ago
    
This fits quite well with the OP's example of infinity times zero. – Owen 3 hours ago

Infinity is a special case. People underestimate the concept of infinity (even those who know higher math!). It is best defined as something unreachable (a limit), and it makes sense that the normal rules break down in weird ways when you try to play with it.

Of course there are people who disagree, but that is beyond the scope of this question

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The ordinal numbers aren't really a matter of agreement or not; they just simply exist. Even people who don't believe in the validity of uncountable sets accept the existence of (countable) ordinals bigger than $\omega$, because you can prove they exist very, very easily, assuming only the existence of $\mathbb{N}$ and a few very basic axioms. – goblin 15 hours ago
    
@goblin I'm not exactly sure what you are saying. Can you provide a link to the proof? – Geeky I 14 hours ago
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I can give you the gist of the proof right here. Given well-ordered sets $\alpha$ and $\beta$, define $\alpha+\beta$ as the set $\{(a,0) : a \in \alpha\} \cup \{(b,1) : b \in \beta\}$. Promote this to a well-ordered set by declaring that $(x,i) \leq (y,j) \iff (i<j) \vee (i = j \,\&\, x \leq y).$ You can check that this is indeed a well-order. Now think about what $\omega+1$ looks like. If you don't trust intuition (which is fair enough), try checking that $\omega+1$ has a greatest element. Ergo, since $\omega$ does not, they cannot be isomorphic. – goblin 14 hours ago
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Notice we haven't really made use of the axiom of powerset. So this is a very "strong" proof. Even mathematicians who reject the existence of cardinalities greater than $\aleph_0$ accept this proof. – goblin 14 hours ago

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