Question

If k is an integer, then k^2+3k is even.

Proof by cases

Answer 1

or is there only two cases, when k is odd or k is even

Answer 2

Well, since we're dealing with parity anyway, your instincts should tell you to consider the case where k is odd, or k is even.

Answer 3

Already got it?

Answer 4

Can you help me out with another?

Answer 5

Will try :)

Answer 6

If a and b are aribitary real numbers, then a*b<= (a^2 + b^2 )/2

Answer 7

I was thinking of starting by saying 0<=a^2+ b^2-2a*b, but i dont know if thats gonna get me anywhwere.

Answer 8

hmm
a² - 2ab + b² = (a-b)²

Answer 9

hmm....

Answer 10

its not coming to me, i dont know what to do with 0<=(a-b)^2

Answer 11

You do agree that
(a-b)² is always nonnegative, right? as it's a square? :)

Answer 12

oh right, i didnt even think about....didnt get enough sleep last night i guess

Answer 13

so, if a and b are real numbers, it's always true that
0 <= (a - b)²

Expanding...

0 <=a² - 2ab + b²

And the rest is history :)

Answer 14

okay, but then from here: 0 <=a² - 2ab + b²

im still lost a little bit, can you explain the concept, from here: 0 <=a² - 2ab + b², how to we aregure that a*b<= (a^2 + b^2 )/2

Answer 15

Well, using the addition property of inequality, we can add 2ab to both sides, giving us
2ab <= a² + b²

Now using the multiplication property of inequality, we can multiply both sides by 1/2, and since it's a positive number, the inequality doesn't change...
ab <= (a² + b²)/2

voila

Answer 16

i think i finally got it, thanks :)

Answer 17

No problem :)

Answer 18

Let x be an arbritary real number. If |x-2|>4, then |x|>2. Prove.