Why 73 is a prime number?

why 73 is a prime number?

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Because it is not divisible by anything other than one and itself
And because you touch yourself at night

73 : 0.5 = 146

It's not divisible by any positive integer besides 1 and itself.

floor(sqrt(73)) = 8
73 % 2 != 0
73 % 3 != 0
73 % 5 != 0
73 % 7 != 0
Hence 73 is prime.

Suppose 73 isn't a prime number. Then
[math] 72! \equiv 0 \mod 73 [/math]. But then it can be checked by Euclid's algorithm that all numbers from 1 to 72 have inverses mod 73 and therefore [math] 1 \equiv 0 \mod 73 [/math]. Which implies that 73 divides 1, a contradiction.

In other words, 73 is prime because if it wasn't then math would be inconsistent and black holes would appear and shit.

Many of numbers you meet will be unoriginal, being merely products of other numbers, and thus many of these numbers' properties can be reduced to those of their constituents. But every once in a while you come across numbers so fundamental that you cannot help but marvel at their permanence and refusal to be divided. These primal numbers are rightly called primes and they have been the object of mathematical fascination since times immemorial. 73 happens to be one of them.

add more to that rule, its too vague to be true.

73/2=36.5

Fight me

>floor(sqrt(73)) = 8
>using logic more complicated than the main question to answer the question, without explaining the logic

When someone says somethins is divisible, it means that the quotient must be an intenger with no remainder.

p is divisible by q if p/q=k , where k is included in the set of integers.

>When youre so autistic that you study primes which have no importance usage or application whatsoever

If you don't understand the most basic concepts about what makes a number prime than you have no business on the internet. You need to get down to the gun shop and grab a pistol; buy it on credit and grab a bullet and suck it down like you daddy's cock

I like yo style, also checkem

can you explain pls?

Because it's absolutely value is only divisible by 1, 0, and itself.

Because 7+3 is 10, which has prime factors 5 and 2. 5+2 is 7, and 5-2 is 3. QED.

When you factor a number, you get something that looks like this:
64: 1, 2, 4, 8, 16, 32, 64
4*16=64
2*32=64
1*64=64
and 8*8=64

If you look, you see that you can essentially "fold" over the numbers around the number 8, that's because sqrt(64)=8. Of course, non perfect square composite numbers like 66 (1, 2, 3, 6, 11, 22, 33, 66) folds between 6 and 11, and sqrt(66) falls between those two. This means that you only need to check up to the square root of a number to check if it's prime. e.g. sqrt(67)~8.1, so we check all the primes before it, 67/2, 67/3, 67/5, 67/7, and none of those are integer values, so we don't have to go further.

This also means that any number below 120, (121 is 11^2) only requires you to check up to 2,3,5,7. I actually had a teacher that said this "a number is prime if it's not divisible by 2,3,5 or 7", I asked "what about 121?", and he got confused. It sounds insane but there are actually people this dumb out there.

Also, the notation in his post (73 % 2 != 0) means that the remainder when you divide 73/2 is not zero (meaning you have something after the decimal)

because the digital root is 1

73 = 8*9 + 1 = 2*2*2*3*3 + 1.
And the product of primes plus 1 is also prime.

it is the boaty prime

>the product of primes plus 1 is also prime
15 = 2*7 + 1
Interesting

3*5+1=16
Are you guys retarded?

Are you?
>the product of primes plus 1
>8*9+1
>implying 8 is prime
>implying his post isn't bait

every number can be expressed as a product of primes + 1, ergo every number is prime

>And the product of primes plus 1 is also prime.

7*3 +1 = 22

hmmmmmmmmm

divisibility only applies to integers

I'm pretty sure he meant the product of the n first primes +1 is always prime

Pretty sure he was shooting for Mersenne Primes
> Mersenne prime is a prime number that is one less than a power of two. That is, it is a prime number of the form Mn = 2n − 1

2^2=4
3^2=9
5^2=25
7^2=49
11^2=121

It is not a product of smaller prime numbers, so the fundamental theorem of arithmetic tells us it is a prime number itself.

because the cyclic group on 73 elements contains no proper nontrivial subgroups

But you can't just list random numbers and reasonably expect a dumbass like op to understand. You have to use fta to show that you only need to check for prime divisibility of a number n up to sqrt (n)

what is cryptography

whoops meant for

Primes do have an importance, they are used in cryptography. Google RSA encryption algorithm.

Jesus Christ p bad teacher desu

Because if you turn 73 you get 37.
3 and 7 are both prime ñumbers.
37 is a prime ñumber.

This is like self-defense instructor saying "you can only attack me like this"

Explain this method

By definition a prime only has integer divisors of one and itself which result in an integer number

Kek

>3 and 7 are both prime
>37 is prime
Huh? Not sure if fucking with OP but that is not the way it works, counter example: 3 and 5. 53 is prime; 35 is not.

Is 0 a prime number?

Why show up to your self defence class with a gun then?

0 has the most divisors out of any number

how do you tell if a complex number is prime?

Look up Gaussian Integers and Gaussian Primes
en.wikipedia.org/wiki/Gaussian_integer

>why 73 is a prime number?
Because is in the Bible

73/0.5= 73 times 2,

You dingus