Make expressions equal to 5 using three 4s puzzle

L

lookagain

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Use three 4s in each expression, but no other digits/numbers, and these other constraints
below to make expressions that are equal to 5.

This puzzle is in base 10.

You may use:
--------------

addition
subtraction
multiplication
division symbols
parentheses
square root symbol or sqrt( ) a finite number of times
factorial sign (regular use only)

You may not use:
-------------------

percent symbol
decimal points
exponentiation
logarithms
trig functions
floor function
ceiling function
concatenation of digits
any other functions
any other characters

I will give an example, and then I am asking for an eventual total of four more expressions
(you can post just one at a time if you would like) using a spoiler please.

4 + 4/4 = 5

Or, in Latex in vertical style, \(\displaystyle \ \ 4 + \dfrac{4}{4} \ = \ 5\)

As you come up with candidates, keep in mind that the following examples, for instance,
would not be considered essentially different solutions:

4/4 + 4 = 5

\(\displaystyle \ \ 4 + \dfrac{\sqrt{4}}{\sqrt{4}} \ = \ 5\)

\(\displaystyle \ \ 4 + \dfrac{4!}{4!} \ = \ 5\)
 
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These expressions are two versions of the same solution:

4 + (√4 - √4)!

4 + (4 - 4)!

?

[imath]\;[/imath]
 
Otis said:
These expressions are two versions of the same solution:

4 + (√4 - √4)!

4 + (4 - 4)!

?

[imath]\;[/imath]
Click to expand...

I would take the second solution as the basic one, whereas the first one could have the symbols removed from
the numbers inside the parentheses, and it would still have the same meaning (as you already indicated to me).
 
lookagain said:
I am (still) looking for one more solution from the solvers.
Click to expand...
Before I spend any more time trying to find the elusive fifth expression, could you please assure me that there is, in fact, a fifth expression?
Don't tell us what it is (yet!), just that there is one more. :)
 
Harry_the_cat said:
Don't tell us what it is (yet!), just that there is one more.
Click to expand...

Hint:

Harry_the_cat, notice how the first solution (my example) and the second listed solution of post #4, the fourth solution, are related.

Now, can you (or someone else) find a related solution to the one given in post #3 for my fifth solution?

------------------------------------------------------------------

Sent on my cell phone
 
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Haven't looked at other solutions, mine are probably all duplicates:
(4! - 4)/4
SQRT(4! + 4/4)
4 + (4-4)!
 
@lookagain, are you say that there are exactly 5 solutions? How can these be shown? I suspect that it can only be shown by looking at every possible case.
 
Steven G said:
@lookagain, are you say that there are exactly 5 solutions? How can these be shown? I suspect that it can only be shown by looking at every possible case.
Click to expand...

These are the ones I have found. If anyone else finds more with the same rules
and posts them, then that person will get extra recognition.
 
(4! - 4)/4
4 + 4/4
4 + (4 - 4)!
√(4! + 4/4)
√(4! + (4-4)!)
A red highlight shows sub-expressions that evaluate to 1
@Otis found the sub-expression x=(√4 - √4)! which might have been used in both "4+x" and "√(4! + x)" but it wasn't sufficiently different to x=(4 - 4)!

How about the slight adjustment x=( (√4)! - √4 )!
 
Cubist said:
(4! - 4)/4
4 + 4/4
4 + (4 - 4)!
√(4! + 4/4)
√(4! + (4-4)!)
A red highlight shows sub-expressions that evaluate to 1
@Otis found the sub-expression x=(√4 - √4)! which might have been used in both "4+x" and "√(4! + x)" but it wasn't sufficiently different to x=(4 - 4)!

How about the slight adjustment x=( (√4)! - √4 )!
Click to expand...

Those would be just more symbols that could be dropped to get it back to a basic solution. In the same way,
I would not consider . . .
\(\displaystyle x \ = \ \dfrac{(\sqrt{4})!}{\sqrt{4}} \ \) to be used as part of a solution.

Also, the factorial sign would not be used on an expression in these puzzles that has a value of 1 or 2, because it does not change the value.
 
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