4 ms·

First day of college: "prove that 0 times x is equal to 0". That got rid of everyone who wasn't really into maths :-) That exercise was literally the first que

First day of college: "prove that 0 times x is equal to 0". That got rid of everyone who wasn't really into maths :-) That exercise was literally the first question, on the first hour of back to school, after 5 minutes of a welcome speech. I can't even figure out the proof anymore, although it really can't be complicated given the ridiculously small amount of tools you are given (basically x + -x is 0, and (a+b) times c equals a times c + b times c. You are given nothing else.

deleted 17y ago
[deleted]
antiform 17y ago
From the real number axioms: Let x \in R. Then we have that x*0 + x*0 = 0*x + 0*x = (0+0)*x = 0*x = x*0 by commutativity of multiplication, distributivity, the existence of a neutral element 0 for addition, and the commutativity of multiplication once again. Thus, we have that x*0 + x*0 = x*0. For clarity, define A = x*0. Then we have that A + A = A A + A + (-A) = A + (-A), by the existence of additive inverses. A + (A + (-A)) = (A + (-A)), by associativity of addition A + 0 = 0, by the definition of additive inverses A = 0, by the definiton of the neutral element 0 of addition which gives us our desired result.