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In general, when given some equation with an arbitrary number of variables of the form

$$f(a,b,c,...)=g(a,b,c,...)$$

we can manipulate the equation and the equality holds true. For example, we can multiply by constants or variables, subtract constants or variables from each side, divide by constants or variables, raise to powers, etc.

Does this hold true for inequalities (non-equality?)? For example,

$$f(a,b,c,...)\neq g(a,b,c,...)$$

Can we still multiply, add, subtract, divide, raise to powers, take logarithms, etc., and the inequality will remain?

YuiTo Cheng
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Hurricane Development
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    Mostly, yes. One main exception would be multiplying by $0$ (there are other considerations, too, such as $-1\neq1$ but $(-1)^2=1^2$). – Clayton Feb 12 '19 at 06:34

2 Answers2

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Consider this: $-1 \neq 1$ but $\left( -1 \right)^2 = \left( 1 \right)^2 = 1$. So, definitely, raising powers does not preserve the $\neq$.

However, addition is just translation. If you translate two distinct point, they remain distinct (in fact, with the distance preserved). So, addition (and hence, subtraction) preserves $\neq$. Similarly, multiplying by a non - zero number is same as scaling the given number. So, distinct numbers remain distinct. Hence, multiplication by a non - zero number (and hence division) preserves $\neq$.

Finally, if you multiply anything by $0$, we get $0$. So, multiplication by $0$ definitely does not preserve $\neq$.

Coming to your last question, exponentiation and logarithms are one - to - one functions. Hence, they must preserve the inequality.

Note: I gave examples of numbers only because $f \left( a, b, c, \cdots \right)$ and $g \left( a, b, c, \cdots \right)$ are actually real numbers, if $f$ and $g$ are real functions. However, if they are complex functions, you may not be able to say the same thing about exponentiation and logarithms as above. Can you think what will happen if we have two non - equal complex numbers and we raise them to $e$ or take their logarithm?

Aniruddha Deshmukh
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Although $\pi\neq0$, we have $\sin\pi=\sin0$. Also, while $a=b$ and $b=c$ always implies $c=a$, the same conclusion cannot be drawn if equality is replaced by inequality. For example, $1+1\neq3$ and $3\neq4-2$, but $1+1$ does equal $4-2$.

John Bentin
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