Question

Prove that for all integers a, a^2 >=a

Answer 1

what method are you supposed to use?

Answer 2

it is certainly true for all negative integers, because \(a^2>0\)

Answer 3

right well its certainly true no matter what but proving it is the tough part for me

Answer 4

i'm pretty sure i'm supposed to split it into 3 cases: a is a natural number, -a is a natural numer, and a=0

Answer 5

and a is a natural number is the one i'm stuck on

Answer 6

for positive integers you could use induction

Answer 7

i've heard that before, but i can't remember what it is

Answer 8

try this
since \(n(n-1)>0\) if both \(n\) and \(n-1\) are, then
\[n^2-n>0\] and so \(n^2>n\)
handle the case 0 and one separately

Answer 9

so this means n>2 though in this example right?

Answer 10

i mean 1(1-1) isn't greater than 0

Answer 11

no problem with 1 since \(1^2=1\)

Answer 12

but i'm just confused as to how you can assume n(n-1)>0

Answer 13

i guess just because you're assuming n>1?

Answer 14

You do not need to assume that n > 1.

For n < 0, it is relatively obvious n < 0 and \(n^{2} > 0\)
For n = 0 or 1, we have \(n^{2} = n\)
The only hard part is for n > 1
There are a few ways to proceed. I like this one.

\(4 = 2^{2} > 2\)

If \(a^{2} > a\), we have

If \(a^{2} + 1 > a + 1\), and then

If \((a+1)^{2} = a^{2} + 2a + 1> a^{2} + 1 > (a + 1)\), and the proof is complete.

Answer 15

that bottom line is a little fuzzy, what are you doing there?

Answer 16

like where's the (a+1)^2 coming from?

Answer 17

The bottom line is the meat. It came from factoring. Algebra 1.

Also, @Satellite73 is right on target. After demonstrating for n < 2, just like I did, n*(n-1) > 0. It is simple and clear.

Answer 18

I know, i understood what he said, and i understand that you factored to get from the first to second step in the bottom row, but then the 2a just disappears...

Answer 19

I didn't get from the first step to the second step. You're reading it backwards. One piece at a time.

\(a^{2} > a\)

\(a^{2} + 1 > a + 1\) -- Just addition

\(a^{2} + 2a + 1 > a^{2} + 1 > a + 1\) -- Just addition

\((a+1)^{2} = a^{2} + 2a + 1 > a^{2} + 1 > (a + 1)\) -- Factoring

It's not magic. I'm just heeding the inequality. Increasing one side by a positive value maintains the inequality. Factoring an expression - not changing the value at all - maintains the inequality.

Answer 20

I see what you did, I didn't realize you were adding a third part to the inequality, thats a really cool way of doing it. and you can do that since you're assuming a>1, right?

Answer 21

Correct. We needed to get rid of the negative numbers (since adding 2a might violate the inequality) and the equalities (because they are annoying).

It's called "Mathematical Induction". We used this: If \(a^{2} > a\), then \((a+1)^{2} > (a+1)\) and proved it by direct construction. Since 'a' is arbitrary, we just proved strick inequality for all Natural Numbers > 1.

Answer 22

Ok cool, thanks so much. that makes a lot of sense now.

Answer 23

And thank you too satellite, you helped a lot as well!