Mathematical symbol for 'slightly greater than'?

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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:

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

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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.

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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

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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 $\epsilon\gt0$.

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0 < (a-b) << b

This depends on your definition of "slightly"

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No. There isn't such a symbol.

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

slightly less than

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

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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.

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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 $\varepsilon$.