Question

how do u prove 2^1/2 + 3^1/2 is not an integer?

Answer 1

1. Add them up.
2. Discover that there's something after the decimal point.
3. Jump up and down yelling "OH YEAH I'VE PROVEN IT HAHAHAHAHAHAHA"
4. Laugh evilly
5. Discover that these steps are practically useless
6. Still read on
7. Be angry
8. Hold your computer
9. Throw it into the street
10. Get arrested by police
11. Go to jail
12. Destroy your future

Answer 2

OK here's the real one
1. Prove that \(3<\sqrt2+\sqrt3\)
2. Prove that \(\sqrt2+\sqrt3<4\)
3. Show that \(3<\sqrt2+\sqrt3<4\)
4. Done

Answer 3

i would laugh if i were in mood :),
thnx for the answer but how can i prove that for example 3 is less than the sum?

Answer 4

Assume the opposite.
\[3\ge\sqrt2+\sqrt3\]\[9\ge(\sqrt2+\sqrt3)^2\]\[9\ge5+2\sqrt6\]\[4\ge2\sqrt6\]\[2\ge\sqrt6\]\[4\ge6\]Therefore \(3\ge\sqrt2+\sqrt3\) is invalid.
Therefore \(3<\sqrt2+\sqrt3\)

Answer 5

brb sorry

Answer 6

tyt its ok

Answer 7

actually proof is done i think, thnx for the answer

Answer 8

no problem :)