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May 6th, 2018, 06:54 AM   #1
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prove it Algebra is not function

hi
please
prove it : y^5-8y-x=0

is not function

with these method : (X1=X2)→(f(X1)=f(X2))
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May 6th, 2018, 09:03 AM   #2
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The values y = ±2^(3/4) and y = 0 all correspond to x = 0.
Hence y is not a function of x if x = 0 is included in the domain of y.
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May 6th, 2018, 09:08 AM   #3
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Thanks,
but I need with this method:

(X1=X2)→(f(X1)=f(X2))

example:
the function

and prove it:

Last edited by skipjack; May 6th, 2018 at 09:11 AM.
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May 6th, 2018, 09:27 AM   #4
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Quote:
Originally Posted by aaaaaaaaaaaaaaaaaa View Post
Thanks,
but I need with this method:

(X1=X2)→(f(X1)=f(X2))
That's what he did. Try to adapt his proof in a form that is suitable for you.
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May 6th, 2018, 09:31 AM   #5
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To employ (X1=X2)→(f(X1)=f(X2)), you need to replace f(X1) with Y1 and f(X2) with Y2 (else you are assuming that the function f exists).

X1 = X2 = 0 doesn't imply that Y1 = Y2, because you could have Y1 = 0 and Y2 = 2^(3/4).

It's important for the above method that (x, y) = (X1, Y1) and (x, y) = (X2, Y2) both satisfy the equation y^5 - 8y - x = 0 that was given.
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May 6th, 2018, 10:26 AM   #6
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Quote:
Originally Posted by skipjack View Post
To employ (X1=X2)→(f(X1)=f(X2)), you need to replace f(X1) with Y1 and f(X2) with Y2 (else you are assuming that the function f exists).

X1 = X2 = 0 doesn't imply that Y1 = Y2, because you could have Y1 = 0 and Y2 = 2^(3/4).

It's important for the above method that (x, y) = (X1, Y1) and (x, y) = (X2, Y2) both satisfy the equation y^5 - 8y - x = 0 that was given.

You are using the value method,
but I need methods without values for Variables
"" just Variable "".

Last edited by skipjack; May 6th, 2018 at 02:16 PM.
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May 6th, 2018, 10:31 AM   #7
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the second example :


and so :
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May 6th, 2018, 03:01 PM   #8
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That example uses a quadratic equation with two distinct roots (for x = 1/S). In my first reply, I supplied the three real roots of your quintic equation that exist for x = 0.

By specifying xS = 1, you excluded the possibility that x = 0, for which your reasoning fails, as +0 and -0 are the same. Thus, you considered (by exclusion) a particular value.
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May 6th, 2018, 04:43 PM   #9
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Quote:
Originally Posted by skipjack View Post
That example uses a quadratic equation with two distinct roots (for x = 1/S). In my first reply, I supplied the three real roots of your quintic equation that exist for x = 0.

By specifying xS = 1, you excluded the possibility that x = 0, for which your reasoning fails, as +0 and -0 are the same. Thus, you considered (by exclusion) a particular value.
EXAMPLE:

x=y-1 , x1=x2 , (y1)-1=(y2)-1 , y1=y2 (i prove that is function)

so :

x=y^5-8y , x1=x2 , (y1)^5-8(y1)=(y2)^5-8(y2) , y1=y2 , y1=?

If I get the second answer for y1 by y2, then prove that is not function
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May 7th, 2018, 02:05 AM   #10
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An equation g(x, y) = 0 defines y as a function of x if and only if the equation has a unique solution for each value of x.

Providing an example where y is a function of x doesn't clarify what you require in regard to another example where it's to be proved that y isn't a function of x.

To prove that y isn't a function of x, you need to show that some value of x exists for which g(x, y) = 0 doesn't have a unique solution (because it has no solution or has more than one solution).
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