Prove by contradiction. Let x, y, and z be positive real number. Prove if x > z and y^2 = xz, then x>y>z.

Question

It deals with proof in mathematics 

Answer 1

 

Given:  x>0; y>0; z>0;  x>z;   y^2 = xz;

 

Prove by contradiction that x>y>z, that is x>y AND y>z

Proof:

 

   Suppose the opposite is true, that is,  x<=y OR y<=z.

  It must be shown that both are false.

 

For x<=y and x>z, then it follows that z < x <= y.

Multiplying everything by y>0 results in yz < xy <= y^2.

                                                          yz < xy <= xz <--- substitution of the given y^2=xz

                                                       So yz < xz by removing the middle inequality

 Dividing both sides by z>0 shows that y < x which contradicts the

 inequality proven above in bold.    So x<=y must be false. Hence x>y.

 

For y<=z and z< x, it follows that y <= z < x.

Multiplying everything by y>0 results in y^2 <= yz < yx.

                                                          xz <= yz < yx.

So again, xz < yx by removing the middle inequality.

After dividing everything by x>0, z<y which contradicts the

statement proven in bold above. So y<=z must be false.

Hence y > z.

 

Therefore x > y and y>z as proven in bold and italic above.

[End of proof]