Two numbers equality formula

Question

Given two real numbers $a$ and $b$, is there any formula which would give $1$ if $a=b$ and $0$ if not ? (I am not talking about conditional expressions). Thanks.

Edit : Everything is allowed.

Answer 1

(It's taken me long enough to write this ... I bet some of these appear in other answers.)

• Using the Iverson bracket: $[a=b]$
• Using the Kronecker delta (not typically used for reals, except when defining discrete distributions): $\delta_{a b}$
• Using contour integration (sneakily): $ \liminf_{n \rightarrow \infty} \frac{1}{2\pi}\int_{0}^{2\pi} \mathrm{e}^{\mathrm{i}(a-b)\theta/n} \,\mathrm{d}\theta$
• Using the indicator function: $\mathbf{1}_{\{a\}}(b)$
• Using a standard not uniformly continuous (on $[0,1]$) sequence of functions: $\lim_{n \rightarrow \infty} (\min\{a/b,b/a\})^n $
• Using the Heaviside step function (in disguise, using the Fourier result that the value at the jump is the mean of the limits from each side): $\left. 4(\frac{\mathrm{d}}{\mathrm{d}x} \max\{x,0\})(1-\frac{\mathrm{d}}{\mathrm{d}x} \max\{x,0\}) \right|_{x = a-b}$
• Using the ceiling function: $1 - \lceil \frac{|a-b|}{|a-b|+1} \rceil$

Edit: Because, of course, one has to get the parity of the last one backwards. Fixed.

Edit: Added limit to contour integral because, sadly, not all pairs of reals have integer differences. And more:

• We can look at the mean of a certain cosine function: $\lim_{A \rightarrow \infty} \frac{1}{2A} \int_{-A}^A \cos((a-b)x) \,\mathrm{d}x$
• We can use the sine integral function: $1-\lim_{N \rightarrow \infty}\frac{2}{N\pi} \sum_{n=0}^N \mathrm{Si}(n \, |a-b|)$
• Iteration: $f^{[1]}(x)=\sqrt{x}$ and $ f^{[n]}(x) = f^{[n-1]} \circ f^{[1]}(x)$. Compute $1-\lim_{n \rightarrow \infty} f^{[n]}(|a-b|)$.

Answer 2

The classic one: $0^{|a-b|}$ ...

Edit: In case you consider $|.|$ as conditional choose $$0^\sqrt{(a-b)^2}$$ instead.

Answer 3

This one would do, if you allow limits:

$$\lim_{n\to\infty} \exp (-n|a-b|)$$

You could also compute

$$2-\mathrm{Card} \;\{a,b\}$$

Or

$$1-\int_{a-b}^{e(a-b)} \frac{\mathrm{d}x}{x}$$

but I don't think it's very clean to rely on $\int_0^0 f(x) \mathrm dx=0$ when $f$ is not defined at $x=0$.

Another limit: taking $\frac{2}{\pi}\arctan (b-a)$ you get a number in $]-1,1[$, hence you can use the limit $\lim_{n\to\infty} (1-x^2)^n$, which is $1$ if $x=0$ and $0$ if $x\in [-1,1]\backslash\{0\}$. All in all:

$$\lim_{n\to\infty} \left[1-\left(\frac{2}{\pi}\arctan (a-b)\right)^2\right]^n$$

Answer 4

You can use this function: $||sgn(b-a)|-1|$

Answer 5

How about this where $\lfloor x\rfloor$ is the floor function of $x$:

$\min\{\left\lfloor 2^x\right\rfloor, \left\lfloor 2^{-x}\right\rfloor\}$.

You can even get rid of the min function by recasting this as

$\frac{\left\lfloor 2^x\right\rfloor+ \left\lfloor 2^{-x}\right\rfloor-\mid \left\lfloor 2^x\right\rfloor- \left\lfloor 2^{-x}\right\rfloor\mid}{2}$

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Edit: As ASKASK correctly pointed out (thanks), the question asked for an indicator of equality between $a$ and $b$ rather than an input of $x$ being $0$. To make the adjustment, $x$ in the post can be replaced by $a-b$.

Answer 6

$\left\lfloor\frac{1}{1+(a-b)^2}\right\rfloor$, where $\lfloor x\rfloor$ - floor function and $a$ and $b$ - real numbers.

Answer 7

$$ \left|\{a\}\cap\{b\}\right| $$

Where $|\cdot|$ denotes the order (number of elements) of a set.

One can come up with many similar examples. For example, $\left|\{(a,b)\}\cap\Delta\right|$, where $\Delta=\{(x,x)\;\colon\;x\in\mathbb R\}\subset\mathbb R\times\mathbb R$.

Answer 8

$\operatorname{sinc}(a\pi-b\pi)$ works for when $a$ and $b$ are integers.

Answer 9

I would use this, but it maybe it is already too conditinal for you:

The indicator function is often used in such cases, so the function takes only two values, $0,1$ in such a way that $$ 1_M(x)=\begin{cases}0,x\text{ is not element of } M \\1,x\text{ is element of } M \end{cases} $$ where M is any set which we choose. So in your case we would take $\{a\}:=M\subset\mathbb{R}$ and get $$ 1_{\{a\}}(x)=\begin{cases}0,x\text{ is not element of } \{a\} \\1,x\text{ is element of } \{a\} \end{cases} $$ meaning $1_{\{a\}}(x)$ equals only $1$ if and only if $x=a$ or in you case $1_{\{a\}}(b)=1$ with $b=a$. Although technically it actually says that it is only an element of an $1$-element set.

Answer 10

Just this: $$ \{a=b\} $$

It evaluates to 1 if the condition inside the curly brackets is true and to 0 if it isn't.