. Let x, y ∈ R \ {0}. Prove that if x < x^(−1) < y...

. Let x, y ∈ R \ {0}. Prove that if x < x^(−1) < y < y^(−1) then x < −1.

Expert Solution

By multiplicative inverse axiom, we have x*x^(-1) = 1

Now observe x < x^(-1), let's take two cases x >0 and x < 0;

a) x > 0 --> Multiply both sides by x, sign of inequality won't change when we multiply with a positive number.

x*x < x*x^(-1) => x^2 < 1 => x belongs to (-1,1) but our domain says x > 0 so x -> (0,1)

b) x < 0 --> Multiply both sides by x, sign of inequality changes when we multiply with a negative number.

x*x > x*x^(-1) => x^2 > 1 => x belongs to (-inf,-1) U (1,inf), but our domain says x < 0 so x -> (-inf,-1)

There will be two range of values of x which will satisfy above relation, x (0,1) and x < -1.

With similar analogy we can say y (0,1) and y < -1

Now these x and y values has to satisfy the relation : x < x^(-1) < y < y^(-1)

Now x < y and x^(-1) < y^(-1) so let's take following cases :

i) x (0,1) and y < -1 first condition will not hold, so discarded.

ii) x(0,1) and y (0,1) if first condition will hold then second won't hold e.g. x = 0.2 and y = 0.5, x < y holds but then x^(-1) = 5 and y^(-1) = 4 so x^(-1) > y^(-1) which contradicts our condition, so discarded.

iii) x< -1 and y(0,1) both conditions will hold because a negative number will always be less than a positive number.

iv) x < -1 and y < -1 again, first condition might hold say x = -3 and y = -2 so x < y but x^(-1) = -0.333 and y^(-1) = -0.5 here x^(-1) > y^(-1) contradicts, so discarded.

Conditions on x and y to satisfy the given relation is : x < -1 and y (0,1).

Colby Messinger answered 1 year ago

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