# Prime number minus 1 is an even number?

Is it true that for every prime number $p$ (except $p = 2$), that $p-1$ is an even number?

I tried it in R (code below) for the first 168 primes (found on wikipedia) and it seems to hold, but I'm not sure if it is always true (maybe its a naturally property of a prime?).

 primes = c(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997) ((primes-1)%%2) == 0

Since from definition of prime: $$\bbox[5pt,border:2px solid blue]{\scriptsize\text{A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.}}$$ It implies that prime number must be odd (except 2) because otherwise (it was even) it would be divisible by 2, so it's predecessor must be even since it iself is odd!
Every prime greater than $2$ is odd because if it was not it would be divisible by $2$. Every odd number minus $1$is even.