像这样的问题很容易通过检查程序的代码并查看程序的状态如何随时间变化来解决。在你的函数print_numbers()（顺便说一句，什么数字？），只有一种状态，即LED的开/关状态。
我在源代码中添加了一些非常冗长的注解。在实践中，在心理上这样做往往就足够了。

def print_numbers():
    while True:                             # loop forever
        board.digital_write(5,1)            # turn LED on
        time.sleep(1)                       # wait for one second
        board.digital_write(5,0)            # turn LED off
        time.sleep(1)                       # wait for one second
        Led_state = board.digital_read(5)   # read on/off state of LED, this will always be off because we turned it off two lines above
        if Led_state == 0:                  # print stuff
            print("led off") 
        else: 
            print("led on")                 
                                            # jump to beginning of loop, start over

使程序打印实际LED状态的一种方法是首先查询，然后将其翻转到其他状态：

def print_numbers():
    while True:
        Led_state = board.digital_read(5)
        if Led_state == 0:
            board.digital_write(5, 1)
            print("led on")                 
        else: 
            board.digital_write(5, 0)
            print("led off") 
        time.sleep(1)

它也可以在不阅读LED状态的情况下完成，保持原始的展开循环：

def print_numbers():
    while True:
        board.digital_write(5, 1)
        print("led on")                 
        time.sleep(1)
        board.digital_write(5, 0)
        print("led off")
        time.sleep(1)