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May 6th, 2018, 06:54 AM  #1 
Newbie Joined: May 2018 From: world Posts: 5 Thanks: 0  prove it Algebra is not function
hi please prove it : y^58yx=0 is not function with these method : (X1=X2)→(f(X1)=f(X2)) 
May 6th, 2018, 09:03 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
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. 
May 6th, 2018, 09:08 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 09:11 AM. 
May 6th, 2018, 09:27 AM  #4 
Senior Member Joined: Oct 2009 Posts: 403 Thanks: 139  
May 6th, 2018, 09:31 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
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. 
May 6th, 2018, 10:26 AM  #6  
Newbie Joined: May 2018 From: world Posts: 5 Thanks: 0  Quote:
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.  
May 6th, 2018, 10:31 AM  #7 
Newbie Joined: May 2018 From: world Posts: 5 Thanks: 0 
the second example : and so : 
May 6th, 2018, 03:01 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
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, 04:43 PM  #9  
Newbie Joined: May 2018 From: world Posts: 5 Thanks: 0  Quote:
x=y1 , x1=x2 , (y1)1=(y2)1 , y1=y2 (i prove that is function) so : x=y^58y , x1=x2 , (y1)^58(y1)=(y2)^58(y2) , y1=y2 , y1=? If I get the second answer for y1 by y2, then prove that is not function  
May 7th, 2018, 02:05 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
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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algebra, function, prove 
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