Question

If n is a prime ,then n+2 and n-2 are prime numbers. Is 19 a counterexample ??? Please ask my question !!!

Answer 1

Answer:

No, if 19 a counterexample, then how?

because n-2=17, yes is a prime no.

but n+2 =21, is a composit no. not prime no.

with an ex. we can take it n=5

because, n-2 =5-2=3, and n+2= 5+2= 7

3 and 7 are also a prime no.

Answer 2

We are told the statement,

"If n is a prime number, then (n - 2) and (n + 2) are prime numbers."

Let the statement break into the following 3 statements:

p : "n is a prime number."

q : "(n - 2) is a prime number."

r : "(n + 2) is a prime number.

Now our statement is in the form $\displaystyle\sf {p\implies(q\land r).}$

On taking n = 19, we get,

p : "19 is a prime number."

q : "17 is a prime number."

r : "21 is a prime number."

Here both p and q are true statements since 19 and 17 are really primes, but r is a false statement since 21 is really composite.

Then,

$\displaystyle\longrightarrow\sf{p=T\quad;\quad q=T\quad;\quad r=F}$

Thus,

$\displaystyle\longrightarrow\sf{q\land r=T\land F=F}$

And,

$\displaystyle\longrightarrow\sf{p=T\ \implies\ \lnot p=F}$

Now consider the form of our statement.

$\displaystyle\longrightarrow\sf{p\implies (q\land r)}$

Since $\displaystyle\sf {(a\implies b)=(\lnot a\lor b),}$

$\displaystyle\longrightarrow\sf{[p\implies (q\land r)]=[\lnot p\lor (q\land r)]}$

$\displaystyle\longrightarrow\sf{[p\implies (q\land r)]=[F\lor F]}$

$\displaystyle\longrightarrow\sf{[p\implies (q\land r)]=F}$

Thus our statement is a false statement. Hence,

"if n is a prime number, then either (n - 2) or (n + 2) is not a prime number in case of n = 19."

Thus one can say that 19 is a counterexample.

But if the statement was like,

"If n is a prime number, then either (n - 2) or (n + 2) is also a prime number."

then 19 would no more be a counterexample.