# Mathematical symbol for 'slightly greater than'?

I am wondering if there is a mathematical symbol which indicates that a value is slightly greater than, or slightly less than, another value. I know there is a symbol which indicates that a value is much greater than, or much less than, another value:

$$a\gg b \qquad\text{or}\qquad a\ll b$$

I am wondering if there is a counterpart to this, which indicates:

$$a\,\text{ is slightly greater than }\, b \qquad\text{or}\qquad a \,\text{ is slightly less than }\, b$$

• Maybe \lesssim $\,\lesssim\,$ or \lessapprox $\,\lessapprox\,$?
– dxiv
Mar 12, 2017 at 6:13
• @dxiv: I would interpret those as "A is either less than B or approximately equal to B" (or equivalently, "A is not substantially greater than B"). Mar 12, 2017 at 8:16
• @Kevin I'd be surprised if either of those were ever used in contexts where $A \gt B$ was even remotely possible. That said, I don't have an authoritative reference handy, which is why I posted it as just a comment.
– dxiv
Mar 12, 2017 at 8:41
• @dxiv In the study of PDEs, for example, $A\lesssim B$ means that there is a constant $C>0$ so that $A\leq CB$. It's meaningless for single numbers $A,B\in\mathbb R$; the point is that the constant $C$ does not depend on various choices (of parameters, functions, or something else). It is not at all rare that $A>B$ and $A\lesssim B$. // I guess almost any symbol works if you add an explanation, but none works without one. Mar 12, 2017 at 11:41
• I always do $$a = b\, (1+\epsilon)$$ with $\epsilon \geq 0$ to indicate that it can be equal or more but in a relative sense. As in 3% more, or 5% more. Mar 12, 2017 at 16:06

## 10 Answers

More often it is used as $b=a+\epsilon$ where $\epsilon$ normally stands for a small positive quantity. That provides b slightly greater than a. Similarly $-\epsilon$ for slightly below.

• While this is certainly conventional, you're still expected to define $\epsilon$ somewhere in the accompanying text (perhaps alongside the definitions of a and b). Otherwise, I have no idea whether you mean "some small real number," or some kind of hyperreal infinitesimal, or even something really exotic like the dual numbers. Mar 12, 2017 at 8:20
• @Kevin the meaning is "some small real number" when it's clear from context, which it is outside of hyperreal analysis or whatnot. Mar 12, 2017 at 18:46
• @Kevin: I would pretty much always assume it means "some small real number" (haven't you heard the famous joke? "let $\epsilon<0$"). But also, I would interpret the other possibilities you mention as formalisms for representing infinitely small quantities anyway, so the conceptual meaning is more-or-less the same (okay, you could take a negative infinitesimal, but why would you?). Mar 12, 2017 at 21:58
• @djechlin I would say it's equally likely in standard analysis that $\epsilon$ is a small positive quantity, or an absolutely small quantity, when not specified or made obvious by its use as eg. a distance.
– jwg
Mar 15, 2017 at 7:52

Perhaps:

$$\lt_\epsilon\quad\gt_\epsilon\quad\lt^\epsilon\quad\gt^\epsilon$$

using the idea that $a\lt^\epsilon b$ means $a+\epsilon=b$, where $\epsilon\gt0$.

• this is really clear for probably just being made up. Mar 12, 2017 at 18:47
• @djechlin All good notation is originally made up by somebody. I haven't seen this before but could see it catching on. Mar 14, 2017 at 12:25
• a<b already implies such an epsilon greater than zero. Law of trichotomy! Mar 14, 2017 at 23:11

No. There isn't such a symbol.

• Perfect answer. Mar 13, 2017 at 13:11
• Goodness, imagine if Euler or Leibniz had had this thought. Mar 13, 2017 at 22:09
• IMO, a very poor and demeaning answer. $\lesssim$ and $\gtrsim$ are used.
– user65203
Mar 14, 2017 at 10:32
• The 3 sentences before the edit may answer the question (though it's hard to be confident when asserting that no symbol exists for a concept, considering the proliferation of symbols...), but the rant following the edit is unproductive and irrelevant. Mar 14, 2017 at 15:07
• @Nye that's.... A bit harsh. I would hesitantly say that the core answer is plausible at a minimum (hard to prove a symbol doesn't exist because there's no standard), but I think the point of this post is to say that there is nothing standard, so one must come up with something. There exist symbols that could be used, but I wouldn't use any without explanation (except $+\epsilon$, which isn't actually a single symbol). Now, I actually think most of the edit is simply a rant and isn't productive, but the first few lines seem fairly accurate, if inelegant and not super precise Mar 14, 2017 at 18:22

The "much greater" symbol $\gg$ is not exactly standard and should always come with an explanation.

There is no standard symbol for "slightly greater". It is the kind of thing best explained in English or made precise mathematically, as suggested by AHusain

• In fact,$\gg$ is rather standard now, as proven by the fact that Latex has a specific symbol for it, recalled by @Henry W. Mar 12, 2017 at 7:35
• @JeanMarie: there is a huge number of symbols in Latex, so it is a rather weak notion of standard. From my meager experience, it seems it is more a physics standard than a maths one. Mar 12, 2017 at 14:26
• As a mathematician I would tend to read '$a \gg b$' as a hypothesis to mean '$a$ is sufficiently large compared to $b$', in other words there is some (potentially extremely fast-growing or even non-constructive) function $f$ such that if $a > f(b)$, then the rest of the argument works. Or it could mean $a$ is unreachable from $b$ in some precise sense, e.g. in nonstandard analysis it could mean $a$ exceeds every finite multiple of $b$. Still, you would have to specify the convention somewhere. Mar 12, 2017 at 15:23
• In analytic number theory, for example, $f(x)\ll g(x)$ has a specific meaning (it's a synonym of $f(x)=O(g(x))$), which doesn't even necessarily imply that $f(x) \le g(x)$. So while the symbol might be standard, there is no one standard meaning for it. Mar 14, 2017 at 6:36
• @Taladris Well, rereading OP's post, he talks about a mathematical symbol, not about a symbol in math. So it is not clear whether he would be satisfied with an answer involving only "math as used by non-mathematicians". In practical engineering there are lots of mathematical symbols used in a way a mathematician would frown at: for example << and >> ("much less/greater than) are typical and their "exact" meaning (insert hand-waving here :-) is context-dependent (and usually understood -by engineers- without further explicit specification). Disclaimer: I'am an engineer :-D Mar 15, 2017 at 20:58

0 < (a-b) << b

This depends on your definition of "slightly"

• If you want to include times where b and/or a are negative perhaps 0<(a-b)<<1 would work better. I suppose this isn't a symbol (which in the case of this question is being used to represent an infix notation operation) but it does represent fairly succinctly what the desired symbol is trying to convey Nov 14, 2018 at 20:07

This could be a comment, but given all the comments under the OP's question, I don't think anyone would notice it. Thus I am posting this as an answer.

I was taught that the symbols for not much less than and not much greater than are $\eqslantless$ and $\eqslantgtr$ ($\eqslantless$ and $\eqslantgtr$) respectively. This notation is a bit old-style, but my teacher said this was what he was taught when younger (which would explain why the notation is, as mentioned before, old-style).

However, it is on a slant, and my teacher did not write it with a slant; instead, he wrote the symbols as $\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\ge}$ and $\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\le}$ (each having a very long typeset command). So, $$a\,\text{ is slightly greater than }\, b \qquad\text{or}\qquad a \,\text{ is slightly less than }\, b$$ is equivalent to $$a\space\,\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\le}b\qquad\text{or}\qquad a\space\,\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\ge}b.$$ But I definitely prefer @dxiv 's comment mentioning $\gtrsim$ and $\lesssim$.

Of course there' s no clear boundary between small,large numbers in mathematics it's possible to describe a given number is slightly greater or less than the other considering very small difference between the two numbers. that is @ infinite Small label.(i,e. by ε ),then b=a+ε or b=a-ε. but only for one number it can be expressed as a- for a number near to left and a+ for a number near to right.

• I think the concept of $\epsilon$ contradicts your first sentence. Apr 19, 2018 at 14:54

I see this question as a bit of fun, so how about this?

EDIT: Here is a more general version that works for both slightly less than and also much less than:

• Your first suggestion doesn't look very natural and has the major drawback of too easily morphing into either $a<b$ or $a = b$. Your second suggestion, on the other hand, has some merit, since it looks like $b$ has just pulled-away from equality with $a$. Mar 14, 2017 at 12:32
• The second of these, using "yi" is redundant. In Chinese, just use "-". Mar 14, 2017 at 23:16
• @richard1941 an unexpected depth I did not foresee... however 'yi' sideways becomes a pipe "|" which would get confused with logical OR. I'm surprised you didn't pick up that the first was 'ba' sideways, which written using arabic notation would cause an infinite problem.
– Vlad
Mar 15, 2017 at 4:19

Some suggestion really no answer:

Greater but approximately only greater than $\approx >$

and

Less but approximately only less than $\approx <.$

Like @Vlad, "I see this question as a bit of fun", so I want to propose something as well.

Since there is no standard notation, this leaves room for creativity.

A lot of people proposed ideas like $<^\varepsilon$ to emphasize the idea that $x$ is slightly inferior to $y$ means that $x<y$, but $x+\varepsilon$ is not.

I want to see the problem the other way around, and propose

$$x=^{<\varepsilon}y.$$

This emphasizes the idea that "$x=y$", which means $x$ and $y$ are almost equals... but $x$ is still inferior to $y$ by a $\varepsilon$.