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This semester, I took a course in real analysis and my proof skills were mediocre at best. Unfortunately, I took to Google for many answers since I could not "start" proofs in the slightest.

A popular type of problem I encountered in the class was proving inequalities. For example, prove that for any $x, y \in \mathbb{R},$ $x\diamond y \leq x^2\diamond y^2$, where $\diamond$ denotes some expression involving the terms I can't come up with right now.

Why is it not valid to do a counterexample of the inequality in the opposite direction? That is, find an example such that $x^2\diamond y^2 \not<x \diamond y.$

My reasoning is, "Well, I can find a real number where the reversed inequality is not true, so it must be true the given direction." But what are my logical fallacies here? Are there parts of the problem I haven't negated? Is it equivalent to prove that there exist $x,y\in \mathbb{R}$ such that $x\diamond y >x^2 \diamond y^2$? (Depending on the context, since I don't even know what $\diamond$ is)

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    $\begingroup$ Showing one counter-example does not prove for all real numbers. $\endgroup$
    – fosho
    Commented Jan 4, 2016 at 9:16
  • $\begingroup$ What do you think of this argument? Let $x$ be a number. Show that $x^2 \geq x$. Answer: The reverse inequality is false for $x = 2$. Therefore the given inequality must be true. $\endgroup$
    – David
    Commented Jan 4, 2016 at 9:17
  • $\begingroup$ Take the inequality $x > x^2$. It seems like you are suggesting that this is always true because $\frac{1}{2} \nleq \frac{1}{4}$. What you have shown is that $x > x^2$ only for the case $x = \frac{1}{2}$. Clearly the inequality is false for $x = 2$ for instance. $\endgroup$
    – Improve
    Commented Jan 4, 2016 at 9:19

2 Answers 2

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Showing one counter-example does not prove that the statement is true for all real numbers.

Consider the statement: when it rains everyone carries an umbrella. If you see one person carrying an umbrella in the rain can you conclude that the statement is true?

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The problem is that you need to prove $x\diamond y \leq x^2\diamond y^2$ for ALL values of x and y. So providing a single example only proves it holds for those particular values.

If it is true for some values and not others, then you could find examples where its true but the statement would still be false because it is about ALL x and y.

Another way to look at it is that you're not taking the negation of the whole statement. the negation of $\forall x, y \in \mathbb{R},$ $x\diamond y \leq x^2\diamond y^2$ is $\exists x, y \in \mathbb{R},$ $x\diamond y > x^2\diamond y^2$ which is a statement that can't be disproven with a counterexample.

Lastly, consider the statement $\forall x,y\in\mathbb{R}$, $xy\leq x+y$. This is clearly a false statement but taking $x=0$ would lead to a "proof" by your reasoning.

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  • $\begingroup$ I'm unfortunately slow at understanding. Here's how I understand now: it is not valid to assume that $x\diamond y > x^2\diamond y^2 \,\Rightarrow\, x \diamond y \leq x^2\diamond y^2$ for one particular value of $x$ and $y$ because even if I can show the opposite direction doesn't work, there could still be numbers in the correct direction where it might not work. Hence, we want to prove it does in fact work in the correct direction for all $x,y \in \mathbb{R}$, which we could show by some ingenious algebraic manipulation. Is this correct or did I miss the point? $\endgroup$
    – Decaf-Math
    Commented Jan 4, 2016 at 9:31
  • $\begingroup$ You should be a bit more careful with phrasing but I think you get the point. Which is that proving a statement holds for 1 value does not mean it holds for all values. $\endgroup$
    – kaiten
    Commented Jan 4, 2016 at 9:36

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