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Problem:

Hi, I'm trying to solve the following proof:

If $a > b \geq 0$,

Prove that $a^2 > b^2$.

But I'm a bit new with proofs and I don't manage to write a proper one. I would be very grateful if someone could explain me which are the steps required to demonstrate the proof. Thank you.

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    You will want to show $a\times a \gt a \times b \ge b \times b$ using the properties of an ordered ring. The first inequality follows from $0 \le a \times (a-b)$ and $0 \not = a \times (a-b)$ – 2017-01-11
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    Hint: if $b – 2017-01-11
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    @Henry Thank you very much! – 2017-01-11

3 Answers 3

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If $a>b$ are positive integers then, multiplying both sides by $a$ gives $a^2>ab$, while multiplying both sides by $b$ gives $ab>b^2$. The transitive property of $> $ then gives $$a^2 > ab > b^2 \Rightarrow a^2>b^2$$

If $b= 0; a> 0$, we have $b^2=0$ but $a^2$ is positive, so we still have $$a^2>b^2$$ Hope it helps.

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    Yes it did, it helped me very much!, thank you! – 2017-01-11
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    @iARn No problem. Happy it helped. – 2017-01-11
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Well, if you want to prove this properly, you need to use the definition of $>$. I will assume we know that there exists a subset $\Bbb P\subseteq\Bbb R$ (the set of real numbers) called the positive numbers which are closed under addition and multiplication, and that $\Bbb R$ satisfies the following:

Trichotomy: For any real number $r\in\Bbb R$, exactly one of the following is true:

  1. $r\in\Bbb P$,
  2. $-r\in\Bbb P$,
  3. $r = 0$.

Now, I will define what $a > b$ means.

Definition: Let $a$ and $b$ be real numbers. We say $a > b$ if and only if $a - b\in\Bbb P$. Write $a\geq b$ if either $a > b$ or $a = b$.

Finally, the proof.

Proof: Suppose that $a > b\geq 0$. I claim that $a > 0$; indeed, $a - b\in\Bbb P$, and either $b - 0 = b\in\Bbb P$ or $b = 0$. If $b = 0$, then $a - b = a$, so $a\in\Bbb P$. If $b\in\Bbb P$, then $(a - b) + b = a\in\Bbb P$, because $\Bbb P$ is closed under addition. Now, I claim that $a > b$ implies that $a^2 > ab$. This follows because $a - b$ and $a$ are both in $\Bbb P$, and so $a(a - b) = a^2 - ab\in\Bbb P$, as $\Bbb P$ is closed under multiplication as well. Similarly, $ab \geq b^2$ (you can verify this yourself, note that you need to consider the case $b = 0$ separately). So now we have $a^2 > ab\geq b^2$. We're almost done. $a^2 - ab\in\Bbb P$ and $ab - b^2\in\Bbb P$ or $ab - b^2 = 0$. If $ab - b^2\in\Bbb P$, then $(a^2 - ab) + (ab - b^2) = a^2 - b^2\in\Bbb P$, so $a^2 > b^2$. If $ab - b^2 = 0$, then $a^2 - b^2 = (a^2 - ab) + (ab - b^2) = a^2 - ab\in\Bbb P$, which is what we wanted to show.

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    Chapter 2 of Numbers and Functions by R. P. Burn has a great set of problems on these inequalities. – 2017-01-11
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    Woah! This is a very enlightening answer, thank you! – 2017-01-11
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$$a^2 > b^2 \Leftrightarrow a^2-b^2>0$$ But $$a^2-b^2=(a-b)(a+b)>0$$ $(a>b\ge0 \Rightarrow a-b>0$ and $a+b>0$)