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How can I prove that:

If $a < b$ then $a^2 < b^2$ for $a,b \in \Bbb N$, using the ring properties on the integers and the following property:

If $m < n$ then exist $d \in \Bbb N$ such that $n = m + d$.

Bill Dubuque
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Angel L
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  • I think the answers below are good, but a small nit-pick: I suspect you probably want to say $m<n$ if and only if there exists $d\in\mathbb{N}$ such that $n=m+d$. this gives a complete characterization, and serves as a rigorous definition, for the relation "$<$" on the natural numbers. without the "if and only if", you do not have a complete definition for $<$, and the problem becomes ill-defined. (indeed, how has the symbol $<$ on $\mathbb{N}$ been defined for you?) note, eg, that Mark Saving's post is implicitly assuming the backwards implication in its last sentence – Atticus Stonestrom Mar 12 '21 at 19:14
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    @AtticusStonestrom Good point. I've noted its importance in an edit. – J.G. Mar 12 '21 at 19:22
  • @J.G. indeed, I just saw :) (+1 to your answer) – Atticus Stonestrom Mar 12 '21 at 19:23

5 Answers5

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Write $b=a+d$ so, for $c\in\Bbb N$, $ac<ac+dc=bc$. In particular, $ab<b^2$. By the same logic, $a^2<ab$. Now verify your definition of $<$ (which is what it is, if you take @AtticusStonestrom's advice) implies $<$ is transitive. Alternatively, note $a^2+D=b^2$ with $D:=d(2a+d)$.

J.G.
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Suppose $a < b$. Take $c$ such that $a + c = b$. Then we have $b^2 = (a + c)^2 = a^2 + (2c + c^2)$. Then $a^2 < b^2$.

Mark Saving
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Use the elementary rules for inequalities, as seen in middle school!

From $a<b$, you deduce instantly that $a^2<ab$ (multiplication by $a>0$).

And also $ab<b^2$ (multiplication by $b>0$).

Last, by transitivity, $\:a^2<ab$ and $ab<b^2$ imply $a^2<b^2$.

Another way: $a<b\:$ means $\:b-a>0$. Multiply by $b+a$ (which is positive) to obtain $$(b-a)(b+a)=b^2-a^2>0\iff a^2<b^2.$$

Bernard
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More straightforward :

  • If $a>b$ and $a,b>0$ (there is no need to take $a,b\in\mathbb Z^{+}$), then you have

$$\ln a> \ln b$$

$$\implies 2 \ln a> 2\ln b$$

$$\implies \ln a^2> \ln b^2$$

$$\implies a^2 >b^2.$$

This is an additional answer that I believe will be helpful to you.

lone student
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Assume $a^2 \ge b^2;$

Then

$a^2-b^2\ge0;$

$(a-b) (a+b) \ge 0$; since $a+b \in \mathbb{Z^+}$

we get by multiplying the two sides of the inequality by $(a+b)^{-1}>0$:

$a-b \ge 0,$ or $a\ge b,$ a contradiction

Peter Szilas
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