How do you find two positive real numbers that differ by 1 and have a product of 1?

1 Answer
Cosmic Defect
Mar 1, 2016

Please check the explanation ...

Explanation:

Assume #x# and #y# to be those positive real numbers such that #x<y#.

Condition 1 : They differ by #1#. That means #y-x=1# ..... (1)
Condition 2: Their product is #1#. That means #x.y=1# ..... (2)

Use 1 (#y=1+x#) to eliminate #y# in 2
#x.(1+x)=1; \qquad => x^2+x-1=0#

Solve this quadratic equation
#x=\frac{-1\pm\sqrt{(1)^2-4(1)(-1)}}{2}#
#x_1= \frac{\sqrt{5}-1}{2}; x_2=-\frac{\sqrt{5}+1}{2}#

Of these two solutions we see that
#x_2 = -\frac{\sqrt{5}+1}{2}; \qquad y_2=1/x_2=-\frac{2}{\sqrt{5}+1}# satisfies both conditions 1 and 2 and so is the solution.

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