My Math Forum prove it Algebra is not function

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 May 6th, 2018, 07:54 AM #1 Newbie   Joined: May 2018 From: world Posts: 5 Thanks: 0 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))
 May 6th, 2018, 10:03 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 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. Thanks from topsquark
 May 6th, 2018, 10:08 AM #3 Newbie   Joined: May 2018 From: world Posts: 5 Thanks: 0 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 10:11 AM.
May 6th, 2018, 10:27 AM   #4
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Quote:
 Originally Posted by aaaaaaaaaaaaaaaaaa 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.

 May 6th, 2018, 10:31 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 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. Thanks from topsquark
May 6th, 2018, 11:26 AM   #6
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Quote:
 Originally Posted by skipjack 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 03:16 PM.

 May 6th, 2018, 11:31 AM #7 Newbie   Joined: May 2018 From: world Posts: 5 Thanks: 0 the second example : and so :
 May 6th, 2018, 04:01 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 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.
May 6th, 2018, 05:43 PM   #9
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Quote:
 Originally Posted by skipjack 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

 May 7th, 2018, 03:05 AM #10 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 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). Thanks from topsquark

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