Prove \(p*q+1 \) is not prime
prime number p is defined as \(1|p\) & \(p|p\) where \(p \in \mathbb {Z} \)
This also implies that a prime number is odd therefore any prime number can be written in the form:
\(2k+1 \) where \(k \in \mathbb {Z} \)
let \(p= 2a+1, q = 2b+1 \) where \(a,b \in \mathbb {Z} \)
\(p*q+1 = (2a+1)*(2b+1) +1 \)
\( (2a+1)*(2b+1) +1 = (4ab+2a+2b+1) +1 \)
\( (4ab+2a+2b+1) +1 = (2(2ab+a+b)+1) +1 \)
\(2ab+a+b \in \mathbb{Z}\) so let \( m = 2ab+a+b \)
\((2m+1) +1\)
this is an odd number + 1 which is an even number.
by deffiniton a prime number is not an even number therefore \(pq+1\) will not be prime.
This looks good! A few comments to perfect your proof:
ReplyDelete(1) Because * has other meanings elsewhere in math, I recommend not using it for multiplication, but rather replacing with \cdot, which will come out looking like \( \cdot \)
(2) Be sure your proof uses complete sentences - the first two lines need a slight edit to be sentences.
(3) Careful about the statement of the definition of a prime number - the definition technically says that a prime is a natural number > 1 such that its *only* divisors are 1 and itself (the > 1 and the only points are missing here).
(4) Note that prime numbers are not all odd, but prime numbers > 2 are necessarily odd (this point needs clarification at a couple of points in the proof)
(5) Rather than repeating lines of the equation, it is more typical to align the equals signs and keep going. You can try using the format
\begin{align}
x &= x \\
&= x
\end{align}
Overall nice work!
PS Also please tag your proof with labels to reflect the method (direct proof) and topics (primes, parity)
PPS I'm sorry, the LaTeX code I used got converted, so you can't see my code. You can see the format for the align environment here: https://latex.wikia.org/wiki/Align_(environment) (note that you can leave one side of the equation blank to continue steps of a calculation)
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