Question

For any real numbers x and y, if x^4 - y^4 = 0 then x=y or x=-y.

Answer 1

true.

Answer 2

It seems obvious, since proof by assume will give us an expected result, however, we need to prove that this is only true for (x,y)∈ℝ.

So, proof by contrapositive?

Answer 3

x^4 - y^4 = 0
=> (x^2+y^2)(x^2-y^2) = 0
=> x^2 = -y^2 OR (x+y)(x-y) = 0

Hmmmm....first root is imaginary stuff?

Second part is clear. x = -y OR x = y.

Answer 4

I think the question is asking how do we prove that
∀((x,y)∈ℝ)|x^4-y^4=0↔{x=y}∪{x=-y}

Answer 5

Oh....yeah. if x and y are real, yes. Clearly.

Answer 6

So the solutions by GT and and mad are partial solutions that would probably pass in a calculus class, but not in analytics.

Answer 7

it is a set of questions preceded with "Prove these propositions:"

Answer 8

this is an intro to proofs class.

Answer 9

Probabilistic proof by assume contra. Solve for n and r such that substituting nx and ry in for x and y for {x,y}≠1?

Answer 10

If the "then" statement is true, there cannot be non-unit values for scalar transforms of x and y such that n and r are real values.

Answer 11

this one i am definitely not following.

Answer 12

Alright, if we know the proposition given, THEN the x=y or x=-y. I'm not sure if that's a universal THEN, but if it is, we need to prove that there are no values x≠y or x≠-y such that this is also true.

Answer 13

okay. but if we do it the way GT did it doesn't come out to a concrete result.

Answer 14

If x^4 - y^4 = 0 then (x^2+y^2) (x^2-y^2) = 0. So that factors to (x^2+y^2) (x+y) (x-y) =0. Therefore, x^2=y^2 and x=y. (x+y) = 0 and (x-y)=0 so x=-y and x=y.

Answer 15

how do you get from
(x^2+y^2) (x+y) (x-y) =0
to
x^2=y^2
?
aren't you supposed to be doing induction in class?

Answer 16

i am just using what i know thus far. induction is still fairly new to me so i wouldn't know how to properly use it in this proof.

Answer 17

ok I got it...

Answer 18

\[x^4-y^4=0\iff x=\pm y\]first step:
check for x=1

Answer 19

Let x = 1 such that 1^4 - y^4 = 0. 1-y^4 = 0. -y^4=-1. ....

Answer 20

try difference of squares...

Answer 21

aright so then (1 - y^2) (1+y^2) = 0 so 1=y -1=y

Answer 22

much better, though note that you could have done difference of squares twice\[(1-y^4)=(1+y^2)(1-y^2)=(1+y^2)(1+y)(1-y)=0\]since 1+y^2>0 we only have to solve the last two sets of parentheses for 0\[y=\pm1=\pm x\]that is a more thorough way to do it ;)

Answer 23

so now we just assume that this is true for some random n\[n^4-y^4=0\]from here out out this statement is gospel
the final step is to show that this implies that it is true for n+1
this can be done by plugging in n+1 for x, then using difference of squares and solving the resulting factored equation
if you do that correctly you will be able to show that y=n+1, which means the statement is true for all n
why don't you try that for a bit?

Answer 24

sorry, the statement is\[n^4-y^4=0\iff y=\pm n\]

Answer 25

you cant stop after showing y = +,- 1 = +, - x?

Answer 26

no because that is just for one particular x (x=1)
we want to show it for all x
if that first statement was enough to prove it for all x I could prove that
1+x=2 for all x
by just doing the first step!
clearly that is not enough...

Answer 27

what is needed is to
1)show this statement is true for some particular value
2)assume that it is true for some random value n (which we have already shown)
3) (this is the most important part) show that those two things /necessarily imply/ that this is true for n+1

that is sufficient to prove the statement

Answer 28

can you help with the result? i'm stuck.

Answer 29

checking x=n+1 we have\[((n+1)^4-y^4)=((n+1)^2+y^2)((n+1)^2-y^2)\]\[=((n+1)^2+y^2)((n+1)+y)((n+1)-y)=0\]again we have that the last two sets of parentheses are the only ones that could possibly be zero, so we have\[n+1-y=0\to y=(n+1)=x\]\[n+1+y=0\to y=-(n+1)=-x\]which I think proves our statement for at least all positive n
I suppose we could have made x=0 (that is obviously true as well)

Answer 30

hm... I'm not sure I see how to prove it for negative numbers, which seems to be required in the problem
sorry, I'll have to think about that part

Answer 31

this has to be done by induction?

Answer 32

I was asking you if you were doing induction in class
no, nothing has to be done that way, it's just a method
it's probably the most common one for proving a statement like that about all numbers though

Answer 33

could this be done just as a simple proof

Answer 34

I'm not sure I know what that means...
a proof requires what it requires

ok here's a real cut-corner 'proof':
you are trying to show that \[x^4-y^4=0\iff y=\pm x\] for all x and y
if you think that just saying\[x^4-y^4=(x^2+y^2)(x+y)(x-y)=0\]the square of any number besides zero is positive, so x^2+y^2>0
therefor by the zero factor property either\[x+y=0\to x=-y\]or\[x-y=0\to x=y\]

Answer 35

scratch the 'if you think that' part*