Question

I need some help with understanding a basic mathematical proof.

Answer 1

Disprove the statement: There is a real number x such that \[x ^{6}+x ^{4}+1 = 2x ^{2}\]

Answer 2

Proof: For all real numbers, x, \[x ^{6}+x ^{4}+1\neq 2x ^{2}\].

Answer 3

Case 1: suppose x = 0, then x^6 + x^4 + 1 = 1 and 2x^2 = 0, so x^6 + x^4 +1 =/= 2x^2.

Answer 4

Case 2 (this is the one I'm having a little trouble understanding) : Suppose \[x \neq0\], then x^2 >0. Since \[x ^{6}=x ^{2^{3}}\], x^6 >0. Therefore x^6+(x^2-1)^2 >x^6, so x^6 +x^4 +1 - 2x >0

Answer 5

(x^2-1)^2 is factored from x^6+x^4+1-2x^2...can I end the proof there?

Answer 6

x is nonzero ---> x^2 is positive

x^4, x^6, etc are all positive as well

Answer 7

and as you pointed out, x^4-2x^2+1 turns into (x^2-1)^2

so (x^2-1)^2 is always positive when x is nonzero

Answer 8

Yes, so x^6+x^4+1 - 2x^2 >0 ...can I say that x^6+x^4+1 >2x^2? Thus x^6+x^4+1 =/= 2x^2

Answer 9

I would do it like this

x^6+x^4+1 - 2x^2 >0

x^6+x^4- 2x^2+1 >0

x^6+(x^2- 1)^2 >0

when x is nonzero, x^6 is positive and (x^2- 1)^2 is positive

so that's why x^6+(x^2- 1)^2 >0 is true and x^6+(x^2- 1)^2 = 0 will never be true

Answer 10

I understand what you did, but how (exactly) does that show x^6+x^4+1 =/= 2x^2?

Answer 11

Or is that supposed to be obvious so I don't have to write anything more?

Answer 12

it shows that x^6+(x^2- 1)^2 =/= 0

Answer 13

which consequently means x^6+x^4+1 - 2x^2 =/= 0

Answer 14

ahh, yes so bringing 2x^2 to the left side gives x^6+x^4+1 =/= 2x^2

Answer 15

yes

Answer 16

thank you so much~

Answer 17

no problem