Python Programming

The comparison operators have the highest precedence, followed by the logical operators, and finally the assignment operator. So you should evaluate the comparison operators first. So z = ((x>3) and (y<1)). y<1 is False, making the whole thing False. Then z is set to False.

The comparison operators have the highest precedence, followed by the logical operators, and finally the assignment operator. So you should evaluate the comparison operators first. So z = ((x>3) or (y<1)). x>3 is True, making the whole thing True whatever the result of y<1. Python actually does not even check y<1 - this is called short-circuiting.

Remember the operator precedence: not > and > or. So this should be parsed as (not x>=5) and (x<3). The first part is False, making the whole expression False.

Always evaluate what is inside any parenthesis first. So x>=5 is True and x<3 is False, making this False. Then apply not which makes it True.

Again, with the operator precedence rules, this should be evaluated as (not ((5+y) > (3*x))) or (4 == (x+3)). The first half is True, making the whole expression True.

Python returns a syntax error, even with just True == not False (try it yourself!) Remember that == has higher precedence than not, so Python is reading this expression as (True == not) False, which does not make any sense. You should write z = True == (not False) or even z = (True == (not False)) if you want to be even more explicit.

This is De Morgan's Theorem. not (A or B) is equal to (not A) and (not B).

If you remember from the previous lesson, you can chain boolean expressions. So 3<x<5 is equivalent to 3<x and x<5.

This one is a trick question. Because of operator precedence, this expression is parsed as (3<x<5==3<x) and (x<5). The first half is a chained boolean expression, so it is equivalent to 3<x and x<5 and 5==3 and 3<x, which is False.

Another trick question. Although there is a parenthesis in the second half, the first half is not grouped together since < and == have the same precedence. So Python assumes that this is a chained expression: 3<x and x<5 and 5==(3<x and (x<5)). Obviously, the last part (5==False) is clearly False.