**You can use any Mathematical expression,any method suitable...stupidity is on.**

**But the challenge is you cannot use any other numbers,you can use 5 and 4 only once and you cannot use '**-**'(minus) sign.**

I know 4 possible methods...Can you find still more ?

No vote yet

1 vote

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## Comments

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TopNewest$\sin{(4+5)}={\lceil}{0.3955...}{\rceil}={\boxed{1}}$

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(5/4)*(4/5)

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uuhgjjjjk

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$\dfrac{\gamma}{4!}$ =$1$

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5%4=1or equivalently 5mod4=1

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$5\quad mod\quad 4,\quad \left\lfloor \frac { 5 }{ 4 } \right\rfloor ,\quad \left\lceil \frac { 4 }{ 5 } \right\rceil ,\quad round\frac { 5 }{ 4 } ,round\frac { 4 }{ 5 } \\ \frac { \Gamma (5) }{ 4! } ,\quad cot(45deg)$

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$\Large\color{#D61F06}{\left\lceil\frac{\zeta(5)}{\zeta(4)}\right\rceil}$

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5%4 = 1 ie., 5 modulus of 4 = 1

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How about .5 repeated plus .4 repeated? Does that break the rule for multiple usages of the digits if the horizontal line notation is used?

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yep..breaks the rule.

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4 mod(5)=1

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$y\quad =\quad 54x\\ then\\ { [\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } ] }!\\ =1$

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You cannot use 2...

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⌊ln(5.4)⌋

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In programming languages (int)5/4=1

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Tan(45)=1

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f\left( x \right)=x\ f(5)-f(4)=1

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I'LL BLOW ALL OF YOUR MINDS!!!! |4-5|=1

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a^0 = 1

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Nope....0 not to be used.

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( 5 / 4 ) * ( 4 / 5 ) or ( 5 / 5 ) / (4 / 4) something like this :D

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Nope...you cant use 5 and 4 two times.

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5% 4

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(5^2+4^2+4)/45

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((5×4)/5)/4 or ((4×5)/4)/5. They work

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Mine is easier 5/4 x 4/5

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$\frac{d(\sigma(\sigma(\sigma(\sigma(\sigma(5))))))}{d(\sigma(\sigma(\sigma(\sigma(\sigma(4))))))} = 1$

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Lawl is that too much... =="

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Where $d(n)$ is the number of divisors of n and $\sigma(n)$ is the sum of divisors of n.

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(4-5)^{2}

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'-' sign should not be used and extra numbers cant be used.

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Is this possible 5 + 4i^2

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This gives 1 but extra 2 cant be used as in i^2.

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Just write it like this: $5+4i \times i=1$

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$i$ while any number other than $5$ or $4$ cannot be used.

Even like that is not correct; used the imaginary numberLog in to reply

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Cot45= 1

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The 3 solution i no is Lint of 4/5 Zint of 5/4 Tan45

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5 = 101 (binary notation) 4 = 100 So. 5 xor 4 = 001

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!0

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$\ln \left({\frac{\exp{(5)}}{\exp{(4)}}}\right)=1$

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Wow....superb !!Gud one👌👌

This one's right.

Edit it to Ln instead of log...

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I wrote it ln then changed it to log, thought that in english countries, log refers to the natural logarithm. Anyway I'm editing it again :)

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$log_e$ is also used.

Yeah...or elseHowever Ln seems better...:)

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5×4=20.Then 20÷20=1

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5 modulo 4

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5%4=1

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5^{0} \times 4^{0}

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Good one but you used 0

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can you please explain the notation....??

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It is 5 raised to 0 divided by 4 raised to 0

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5/4

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5 modules 4 = 1

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5/4=1

Solution : int a=5,b=4,c; c=a/b; (hence c=1)

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Ya....and even c=5%4

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5 \times 4 / 5\times 4

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5 and 4 can be used only once...

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(5

4)/(54)Log in to reply

5%4 = 1

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Write a comment or ask a question... 5!/(5*4!)=1

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No two times 5

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5 x 4 -19

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nvm... Can't have a - sign

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Using Exclusive OR gate. 5=101(b) 4=100(b) In Ex-OR logic output will high if input level is different. Here after computation output is 001(b) its decimal is 1

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can i use concantenation?

$\frac{(5|4)}{54} = \boxed{1}$

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You can use concatenations

But two times 5 and 4 not allowed.

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tan 45

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[5/4]

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signum(45),or anything which invlove 4 and 5(but real)

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yes...signum function is possible...It is a function which always gives 1 for any constant.

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Hi Vinay this discussion is a good food for thought....Next

Try to express 0 to 9 numerals using the digit 3 only. One hint is using factorial may come in handy...

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Also post its link in this note as a comment...

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Here, I've created a note for this.

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Thank you...

Generating 0 to 100 will be more interesting..I think you should post a note on it...

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5 modulo 4=1

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(cos 5) ^2 + (sin 5) ^2 = 1

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2 cannot be used.

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We can use 5 XNOR 4 == 1 101 XNOR 100 == 1 What you say?

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5 xnor 4 = $\overline{101\bigoplus100} = \overline{001}=110$

But 5 xor 1 = 001.

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Sorry my fault, It's not a valid solution. Solution should be like, we can not only use 5 and 4? ryte?

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(5+4)/(5+4)

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you've broken the rules

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sum to infinity of i=1 (4*5^-i) //sorry but i don't know how to write the notation

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$\sum_{i=1}^{\infty}4×5^{-i}$

The expression gives the value 1 but extra

1 and $\infty$ cannot be used.Log in to reply

5%4=1 what say guys ?

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5%4

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5%4.

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5%4 (mod)=1

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Taking (5+4)- (4+4) = 1

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54/54 =1,45/45=1,5%4=1,(5*5)%4!

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5^4

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Dim x as integer; x=5/4

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(1) 5%4

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ceil(4/5),floor(5/4)

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gcd (4,5)=1

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what does ceil and floor mean

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(5*4)^0 =1

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0 not allowed and

5 and 4 not allowed twice.

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(5

4)/(54)Log in to reply

5%4

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(4/5)X(5/4)

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Mod(5,4)

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Gcd(5,4)=1 The easiest one

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54/54

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floor(ln(5)/ln(4)) and ceil(ln(4)/ln(5)) I hope ln(x), the natural logarithm, is acceptable here.

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abs(5/4)

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Why not 5 mod 4

p^2(Actually p^2 is Piyush Patnaik)

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5#/4!

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$v_4(5!) = 1$

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Speechless.......

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What is $\nu$ function ???

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Highest power of a prime that divides something.In this case it is not used properly, because 4 is not a prime.

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$v_5(\lceil \sqrt{4!} \,\rceil) = 1.$

I believe the function can still be used... This also works:Log in to reply

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1= 5!/(5×4!)

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5 and 4 only once...

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5-4 floor(5/4) ceiling(5/4) 5

4/54Log in to reply

(5×4)/(10×2)

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10 and 2 cant be used.

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No offence to maths but 4 = IV and 5 = V in roman numerals. So IV/V = I ( after cancelling V on both sides) = 1

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The most obvious $gcd(5,4)=1$

Also, $\frac{\varphi(5)}{4}=1$

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yep..The gcd(5,4) is the most obvious.

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(factorial(5)/factorial(4))/5

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$\frac{\phi(5)}{4} = 1$

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On a similar note of numerical functions, $d(4)-d(5)=1$ where $d(n)=$the number of positive divisors of n.

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yup...another solution..

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*5 mod 4 =1; *(5+4) mod 4=1;

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5 (mod 4) is possible but (5+4)(mod 4) is not possible...only one time 5...

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oh....kk sry

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$\gcd(4,5)=1$

$|cis(45°)|=1$

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yeah....cool solution...gud one ✌👍

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5%4 = 1 :P .. The % is the modulus sign.. It gives the remainder of 2 non- floating numbers ! As the remainder for floating numbers is zero always!

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tan(45)=1

integer(5/4)=1 (we use in C language int(a/b) to get integer quotient)

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The four solutions that i know are:

Tan(45)=15%4=1which is same asremainder when 5÷4where $\Im$ is imaginary number.$\text{FIVE in Roman = V;}$ $\text{FOUR in Roman = IV.So,}$ $\frac{4}{5}=\frac{IV}{V}=I=1$

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$\frac{d}{dx}(x+54)=1$

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5 (mod 4) = 1

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5-4=1

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$5^{\mu(4)}$ is also posible, where $\mu$ defines the MOBIUS function.

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Yes...thats right👍👍 Good use of MOBIUS function

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even cot(45)=1

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Woahh, allaww

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even cotan(45)=1

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yup..right.👍👍

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5 mod 4 =1

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5%4 =1 , 5$\oplus$4 = 1, $\lfloor\frac {5}{4} \rfloor$ =1, $\lceil \frac {4}{5} \rceil$ =1 ,

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-(4-5)=1 The mirror effect

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No minus sign.........

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$5 \oplus 4 = 1$

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You need to describe the modulo number i.e.$5 \bigoplus_84$ for which an extra 8 is required.

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It's not modulo, but an XOR operation, :)

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Whats the sign between 5 and 4???

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this made my day....ahahahah

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5^0 times 4^0 = 1

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Wrong answer 0 can't be used

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nope...0 cant be used.

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totient function of 4 x .5

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here

.5 is not possible...0Log in to reply

small omega of 5 / small omega of 4 ( note small omega counts the number of distinct prime factors )

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totient funtion of 5 / 4

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yeah...its possible

(Totient of 5)/4

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$\dfrac { Gamma(5) }{ 4! } =1$

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Can you please tell me what is it's actual meaning? Like 8! means 8(8-1)(8-2)......(8-7) so what is the meaning of -2! ?? @Michael Mendrin

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Could u explain what the gamma() function's purpose

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The factorial $n!$ is originally defined only for positive integers $n$, so 18th century mathematicians came up with an integral $Gamma(t)=\Gamma (t)=\int _{ 0 }^{ \infty }{ { x }^{ t-1 }{ { e } }^{ -x }dx }$ that has properties similar to the factorial, but defined for all real and complex numbers $t$ (except $0$ and negative integers). It works out that $\Gamma (t)=(t-1)!$. A very versatile function, the gamma function is one of the most widely used functions in mathematics.

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yeah...never thought of this...👌👍

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Aw yes. :-)

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Same thing as 5-4 but looks different $\int _{ 4 }^{ 5 }{ dx } = 1$ Lol. X'D

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yess....even this is possible👍👍

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Nicely done, didn't see that.

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$\lfloor{\frac {5}{4}}\rfloor$ or $\lceil{\frac {4}{5}}\rceil$

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5 and 4 are used twice

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No, they're not. They are used only once in each expression. I have written them separately.

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In a similar way, $\lfloor \sqrt[5]{4} \rfloor, \lfloor \sqrt[4]{5} \rfloor$ etc will also work.

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Or $\lfloor{\sqrt{\frac {5}{4}}}\rfloor$ and $\lceil{\sqrt{\frac {4}{5}}}\rceil$

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Thas possible...greatest integer function..👍

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5/4 remainder is 1 5mod4 is 1

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5%4=1...Yes,thats one of the ways that I knew.

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(5/4) × (4/5) = 1.25 x 0.8 = 1 QED

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The condition is you cannot use 5 and 4 more than once.

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Ah fair enough

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Log54 to the base 54

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5 and 4 can be used only once..

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(5 xor 4) in binary

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Greatest integer(5/4)

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Yes..this one is possible.

3 more ways are possible.

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Signum(5+4), Signum(5×4) Signum(5÷4)

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Again,find the other ways...

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