Is It Obvious?

Assume that $a$ and $b$ are positive with $a > b$ .

If $c > 0$ , then $a^c > b^c$ .

However, if $c < 0$ , then $a^c > b^c$ is false!

In mathematics, if we say a statement is "true", then that means it is always true! But if there is even just one case where the statement is false, then the answer is false.

Simple counter example: $a=3 \ , \ b=2 \ , \ c=-1$ . Thus ,

$\large a^c = 3^{-1} =\dfrac{1}{3} < \dfrac{1}{2} = 2^{-1} = b^c \Rightarrow a^c<b^c$

I think it is interesting that everyone is discussing the case where C < 0, but not where C>0 and a, b < 0. In this case, if C is even, a^c < b^c. Ex: a=-1, b=-3, c=2. a is greater than b, a,b, and c are non-zero reals, so the parameters of the question are satisfied, but a^c=1, and b^c=9, so b^c>a^c.

A note to those who haven't taken math since high school. "TRUE sometimes" in mathematics is an automatic "FALSE". in math "TRUE" is automatically short for "TRUE in all cases"

If $c$ is negative then this inequality is False because $a>b \implies \frac{1}{a}<\frac{1}{b}$

If it's negative it doesn't work. Negative... A^-1=1/A^1 If you got it wrong, it works with c=positive integers only and if a and b=Positive integer(unless c is even), It s sometimes true, not always

For all the Positive integer this statement , when a>b ; $a^{2} > b^{2}$ is true. but in case of negative number this Statement is false as we know every sqrt gives us two root values, one of which is negative. Like a = -2 and b= -3 , as always here a>b. But their squared value $a^{2}$ = 4 and $b^{2}$ =9 ,thus $a^{2}$ < $b^{2}$ .

True/False questions are the easiest to answer by attempting to produce a counter-example, i.e. a contradiction. If a = -1, b=-2, and c = 2, for example, you would have

(-1)^2>(-2)^2 1>4, which is a contradiction. Therefore, it's false.

The question clearly tells that a and b are non zero reals. So they can even be negative reals. let a and b be -1 & -2 respectively. Let c be -3. Then you will get why the answer is false

Answer is false, we can do this problem using proof by counter example. Since we're able to use non-zero real numbers, we can use any number that is not zero, for example (½) > (¼) if we let c = -1, the inequality becomes (½)^-1 > (¼)^-1 , which would be 2 > 4 which is never true in any situation.

The 'then' may be seen as a logical implication. Such an implication only exists if the subsequent statement (that a^c > b^c) is true whenever the preceding statement (the conjunction of the two criteria given in the question) is true. Because this is not the case, the implication does not exist. Whilst some may see this question as ambiguously written, it is simple to recognise that a^c < b^c can not be concluded from the criteria. Furthermore, for those who argue that the question needed to be specific about sign, it was: 'non-zero reals' incorporates all positive and negative real numbers, except zero.

There's no problems with the question at all. There's no need of an "always true" or "always false". If [conditions] then [conclusion]. If there is a conceivable scenario where the conditions are met yet there's another conclusion, then the statement is FALSE.

In one condition where a is 1 and b is -1 and c is any even number then a^c=b^c

The condition becomes false when c is a odd negative integer.

Counter example is one way to conclude the truth value of a statement..but doing in general where there are few cases (a,b>0 and c>0; a,b<0 and c>0; a,b>0 and c<0; a,b<0 and c<0; and so on...) certainly gives idea whether the statement is completely false or satisfiable for certain conditions..just to give a better perspective.. I do know that if we can give any counterexample to negate the statement, that would be the easiest way..

if a = -1, b = -2 and c = 1, then a>b. However if c = 2, b>a.

It depends in c whether the functions would be increasing or decreasing.Inequality won't hold for decreasing functions.

This is trick question. Only limitation on C is it being non-zero, it could be anything else meaning there are an infinite possibilities where the answer is false... But if you assume C is a positive integer then you would get this wrong hence it being a trick question.

Some example : 1) If a = 2, b = 1, c = -3 => a^c = 1/8 < 1 = b^c.

2) If a = -1, b = -2, c = 2 => a^c = 1 < 4 = b^c