Show the rule of divisibility for 2

The rule of the divisibility for 2 says that if a number is divisible by 2, the last digit of the number must be divisible by 2.

For example, 3468, the last digit of the number is 8. Hence, 3468 is divisible by 2 without leaving remainder.

Then how to prove the rule in general form?

Let’s first evaluate 3-digit number that is denoted as \overline{abc}, in which a, b, c are three digits at hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \overline{abc} is rewritten as

\overline{abc}=100a+10b+c

Divide both sides by 2

\dfrac{\overline{abc}}{2} =\dfrac{100a+10b}{2}+ \dfrac{c}{2}

=50a+5b+\dfrac{c}{2}

The equation shows that if \overline{abc} is divisible by 2 if and only if \dfrac{c}{2} is divisible by 2, that is, c must be an even number.

In a more general form, any integers can be expressed as

a=a_n10^n+a_{n-1}10^{n-1}+\dots+a_110+a_0

=10\cdotp a_n 10^{n-1}+10\cdotp a_{n-1} 10^{n-2}+\dots+10a_1+a_0

Divide the equation by 2

\dfrac{a}{2}=\dfrac{10\cdotp a_n10^{n-1}+10\cdotp a_{n-1}10^{n-2}+\dots+10a_1}{10}+\dfrac{a_0}{2}

=5\cdotp a_n10^{n-1}+5\cdotp a_{n-1}10^{n-2}+\dots+5a_1+\dfrac{a_0}{2}

which shows, a is divisible by 2 if and only if a_0 is divisible by 2, that is , a_0 is an even number.