Make 1 with any 3 of the same numbers

Question

Is it possible to make 1 with any 3 numbers, all the same, such as using 3, 3 and 3 or 6, 6 and 6? Can you think of a proof which will work no matter which number it is? Trig functions allowed.

Answer 1

It is possible.

$x^{x-x} = 1$ for all $x \in \mathbb{C}$

See it in action here.

Answer 2

There are certainly many ways. Here are a few ones.

$$\cos\left(x\left(x-x\right)\right) = \cos\left(0\right) = 1$$

or

$$\sec\left(x\left(x-x\right)\right) = \sec\left(0\right) = 1$$

or

$$\left(x\left(x-x\right)\right)! = 0! = 1$$

These all rely on the fact that $x\left(x-x\right)$ is 0 for every $x\in\mathbb C$.

Additionally (and this could be done with only one copy of any complex number $x$),

$$\lfloor\left|\sin\left(\left|x+x-x\right|\right)\right|\rfloor!=\lfloor\left|\sin\left(\left|x\right|\right)\right|\rfloor!=1$$

Where $\left|x\right|$ is the modulus of $x$ and $\lfloor r\rfloor$ is the value of the floor function at $r$, where $r\in\left[0,1\right]$ for any $x$. The same works if $\sin$ is replaced with $\cos$ or $\lfloor r\rfloor$ with $\lceil r\rceil$.

Finally, if we are allowed to cheat and use set theoretical constructs, then

$$\left|\{x\}\cup\{x\}\cup\{x\}\right|=\left|\{x\}\cap\{x\}\cap\{x\}\right|=\left|\{x\}\right|=1$$

will work for whatever (number) $x$: the cardinality of a set containing only one element is 1. This is true even if $x$ is not a complex quantity.

Answer 3

For $x \neq 0$:

$$\left(\dfrac{x}{x}\right)^x=1^x=1$$

Alternatively, the following works for all $x$:

$$\left(\dfrac{\cos(\sin(x))}{\cos(\sin(x))}\right)^x=1^x=1$$

Answer 4

Here's one that works for 0:

$x!^{x - x} \forall x \in \{0, 1, 2, 3, ...\}$ (adapted from Zerris' answer)

Answer 5

From a more computer science angle (depending on how your language handles boolean to integer conversions):

(int)(A == (A & A))
Means: ConvertToInteger(Is A equal to (A bitwise-anded with A)?)
Because: Bitwise-anding A with A always gives A, and that A always equals A, so the result is True, which typically evaluates to 1.

Probably only works for integers, though technically it would work for floats if you had a language that allowed it.

Answer 6

This should work too:

$x^{x-x}=x^0=1$

Answer 7

This works, too, but it's in the category of trick question not mathematics

$1 - 1 + 1 = 1$

also, for $x \ne 0$ this works:

$x^0 - x^0 + x^0 = 1$

working the problem almost suggests a possible quadratic form, if $a$ and $b$ are each $1$, or as above if each evaluate to $1$, and if $x$ is also $1$, as long as $c$ is negative:

$ax^2 + bx - 1 = 1$

this works even for $x = 0$, but I still want the answer to be funny somehow not functional:

$1^x - 1^x + 1^x = 1$

here's a humorous answer, where $x = 6$:

why is $6$ afraid of $7$? because $((x + 1) - (x + 2)) ^ {(x + 3)}$.

($6$ is afraid of $7$ because $7, 8, 9$)

evaluated: because $(7 - 8) ^ 9$. or: because $(-1) ^ 9 = -1$

$-1$ to the power of any number is still $1$, positive if even exponent and negative if odd -- is $-1$ to the power of $0$ a positive $1$? such obscure theoretical math, is there even a right answer or did the first person ever to write a book about math just decide which sign she wanted to attribute to the unusual $(-1) ^ 0$ case?

Another way to phrase the punchline to the joke could be:

"because $-1$ is the loneliest number."

since $(-1) ^ 9 = -1$ you get the "because $7, 8, 9$" answer as the lazy student's "obvious" funny answer that appears to satisfy the problem, but the "right" funny answer expected by the puzzle could be to finish working the problem, arriving at the number $-1$, which would be most clearly suggested as the intended solution like this:

"Why is $x$ afraid of $7$? Because $((x + 1) - (x + 2)) ^ {(x + 3)}$ is the loneliest number."

solve for $x$. what is the loneliest number?

most students will know to find an $x$ that results in "$1$ is the loneliest number" so when they see $x = 6$ it will cause confusion and humor on the way to the intended "right" funny answer that says being negative AND being alone is why $6$ is afraid of $7$.

and as a bonus, the math teacher gets to joke about how NEGATIVE $1$ is surely a lonelier number than POSITIVE $1$, just ask any lonely math teacher!

This is fun stuff for puzzles. Thanks for posing the question!

Answer 8

It is possible and simple.

$(\frac11 )\times1 =1$

$1\times1\times1 = 1$

$x^{(x-x)} =1$ (except $x$ not $0$)