Can someone check pf: If y element reals and (y+1)/(y-2)=x, then x≠1

Thread starter IntroAnalysis
Start date Sep 9, 2011
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In summary, "pf" stands for "proof" in the given statement, which is asking for a verification of the equation. It is important to check if x≠1 in order to ensure that the equation is valid and has a solution. The statement "y element reals" indicates that y must be a real number for the equation to make sense and have a solution. The equation can be rearranged to solve for y, with the values of x and y needing to fall within a certain range for the equation to be true.

Sep 9, 2011

IntroAnalysis

Homework Statement

If y element of Reals and (y+1)/y-2) = x, then x ≠ 1.

Homework Equations

The Attempt at a Solution

Proof by contradiction. Assume x = 1. Then y + 1 = y -2 or 1 = -2. This is false. It contradicts (y+1)/(y-2)= x. Therefore, x ≠ 1.

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Sep 9, 2011

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Related to Can someone check pf: If y element reals and (y+1)/(y-2)=x, then x≠1

1. What does "pf" stand for in the given statement?

"pf" is an abbreviation for "proof." In this statement, it is asking for a proof or verification of the given equation.

2. Why is it important to check if x≠1?

The given equation, (y+1)/(y-2)=x, is undefined when y-2 equals 0, which would make x undefined. Therefore, to ensure that the equation is valid and has a solution, it is important to check if x≠1.

3. How do we know that y must be a real number in this equation?

The statement "y element reals" means that y is an element or member of the set of real numbers. Therefore, y must be a real number in order for the given equation to make sense and have a solution.

4. Can we rearrange the equation to solve for y instead of x?

Yes, we can rearrange the equation to solve for y. First, we multiply both sides by y-2 to get y+1=x(y-2). Then, we distribute the x to get y+1=xy-2x. Next, we subtract xy from both sides to get y-xy+1=-2x. Finally, we factor out y to get y(1-x)=-2x-1 and solve for y by dividing both sides by (1-x), giving us y=(-2x-1)/(1-x).

5. Is there a specific range of values for x and y that would make the equation true?

Yes, there are certain ranges of values for x and y that would make the equation true. For example, if x=2 and y=3, then (3+1)/(3-2)=2, which satisfies the equation. However, there are also values for x and y that would make the equation false, such as x=1 and y=2, which would make the equation undefined. Therefore, it is important to check the given values and make sure they are within the valid range for the equation to be true.