Figure 1 shows the network address 192.168.1.0/24 with the first three octets of the address in decimal and the last octet in binary. The address is over the subnet mask. The first three octets of the subnet mask are in decimal and the last octet is in binary. The host portion of both is all zeros. Figure 2 shows borrowing the first bit in the fourth octet. Figure 3 shows the two subnets one bit borrowed would create. Figure 4 shows that the borrowed bit becomes a one bit in the subnet mask. Figure 5 shows that the two subnets are 192.168.1.0/25 and 192.168.1.128/25 with the new subnet mask of 255.255.255.128.