Proving Incommensurability in Number Theory with a Graph

[e2m2a]

Does anyone know a theorem in number theory or mathematics that could be used to prove the following problem: Given a function f(x) = x sq. Graph the function and with two vertical lines, divide the area under the graph such that the two areas are equal. Denote the point on the x-axis where the first vertical line intersects as point a. Denote the point on the x-axis where the second vertical line intersects as point b. Denote point c on the x-axis as the maximum number in the domain of the function. Thus, the area above the line segment from 0 to a is equal to the area above the line segment from b to c. Prove that the line segment from 0 to a is always incommensurable with the line segment from b to c.

 

[HallsofIvy]

Does anyone know a theorem in number theory or mathematics that could be used to prove the following problem: Given a function f(x) = x sq.

What do you mean by 'x sq." ? x^2?

Graph the function and with two vertical lines, divide the area under the graph such that the two areas are equal.

What "area under the graph". If you meant y= x^2, that graph does not a have an upper bound. Do you you mean the area of the regions having y= x^2 as upper edge, for x> 0 and x< 0? I am going to assume you mean to find two numbers, a and b, such that the area of the region bounded by y= x^2, y= 0, and x= a is the same as the area of the region bounded by y= x^2, y= 0, x= a, and x= b.

Denote the point on the x-axis where the first vertical line intersects as point a. Denote the point on the x-axis where the second vertical line intersects as point b. Denote point c on the x-axis as the maximum number in the domain of the function.

There is no such maximum for y= x^2. And is c a point or a number?

Thus, the area above the line segment from 0 to a is equal to the area above the line segment from b to c.

??Didn't you just say that that was true for y= 0?

Prove that the line segment from 0 to a is always incommensurable with the line segment from b to c.