I don't understand why $2, 147, 483, 647$ is the max number for a $32$-bit integer.
$8$ bits $= 1$ byte$32$ bits $= 4$ bytes
How is this calculated? $8^{32}$ is way over $2$ billion.
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$\begingroup$Bits are the smallest unit of binary number system (like ones in decimal system). So if we have a 32-bit binary number, the largest possible binary number that can be written is $$(11111111111111111111111111111111)_2$$ which can be converted to decimal by the sum $$\sum_{i = 0}^{31}2^i = 2^{32}-1 = 4294967295$$
However, $2147483647$ is the largest integer in Two's Complement (signed in other words), which is represented by $$(01111111111111111111111111111111)_2$$ since first bit is called the sign bit.
$\endgroup$ $\begingroup$A $32$ bit integer can be represented as $b_1b_2b_3\cdots b_{32}$, where all of these are bits (so they are either $0$ or $1$). There are $2^{32}$ possibilities for such integers.
If they are unsigned (i.e. always nonnegative), one can represent the integers $$0,1,\cdots,2^{32}-1.$$ If they are signed, one represents $$-2^{31},-\left(2^{31}-1\right),\cdots,-1,0,1,\cdots,2^{31}-2,2^{31}-1.$$
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