Square an inequality if both sides are positive

[zorro]

I know that we can square an inequality if both sides are positive.
But can we cube an inequality provided both the sides are positive?
If no then why?

 

[Office_Shredder]

The question of whether you can perform an operation on both sides of an inequality is one of whether the function is increasing or decreasing. the function f(x)=x² is not increasing (when x is negative), so will not preserve order.

On the other hand, for just the positive set of numbers, f(x)=x² is increasing, so f(x)<f(y) if and only if x<y.

The function f(x)=x³ is increasing everywhere, so x³<y³ if and only if x<y This means that you can cube both sides of an inequality whenever you want

 

[zorro]

Suppose a<c and b<c (all are positive)
and a,b,c are related by the equation a^2 + b^2 =c^2
I proceeded in 2 ways-
1) a^3 < c^3
and b^3 <c^3
i.e. (a^3 + b^3 )/2 < c^3

2) a^3=a.a^2<c.a^2
similarly b.b^2<c.b^2
i.e. a^3 + b^3 < c^3 (a^2 + b^2=c^2)

why are 1 and 2 different?

 

[Office_Shredder]

1 and 2 are different because you did different things. When you manipulate an inequality you aren't guaranteed to get the best possible new inequality; in this case the first one just happens to be worse than the second one.

Notice you didn't use the fact about the sum of a and b squared when doing (1), which means that it's true for any choices of a, b and c, whereas in (2) a, b and c need to satisfy a²+b²=c²

 

[HallsofIvy]

If 0< x< y, then, multiplying both sides by the positive number x, 0&lt; x^2&lt; xy. Multiplying 0< x< y by the positive number y, 0&lt; xy&lt; y^2. Since "<" is "transitive", 0&lt; x^2&lt; xy&lt; y^2 so x^2&lt; y^2.

Do that again to get cubes.