0

My maths book says that $y^2 > a^2$ implies that $y > a$ or $y < -a$

and that $y^2 < a^2$ implies that $-a < y < a$

Please can someone explain this to me as I do not understand this. Thanks.

D. G.
  • 330
  • 1
  • 8
mku
  • 129
  • Note that $y^2 > a^2$ implies $|y| > |a|$. Then you can split into cases based on the signs of $y$ and $a$. For example if $y$ is negative and $a$ is positive, then $-y > a$, which rearranges to $y < -a$, one of the options your maths book gives you. – D. G. Aug 08 '18 at 16:16
  • @D.G.: $a$ may not be negative. –  Aug 08 '18 at 16:26
  • @YvesDaoust $a$ also may not be positive. The content is symmetric and depends on the sign of $a$ in the same way that it depends on the sign of $y$. – D. G. Aug 08 '18 at 16:53
  • @D.G.: not at all, $y^2<(-1)^2$ does not imply $1<y<-1$. –  Aug 08 '18 at 16:55
  • @YvesDaoust Yes, in other words $y^2 < a^2$ does not imply $-a < y < a$. I think it is clear that the question (and perhaps the asker) lacks the $a \ge 0$ assumption that these results require. – D. G. Aug 08 '18 at 17:03

2 Answers2

0

Because $$0<y^2-a^2=(y-a)(y+a).$$ Now, we can assume that $a\geq0$ and it's obvious enough:

We have or $y+a>0$ and $y-a>0$, which gives $y>a$ or

$y+a<0$ and $y-a<0$, which gives $y<-a$.

The second statement we can get by the similar way.

Michael Rozenberg
  • 194,933
0

The square operation disregards the sign and only deals with the magnitude.

We take for granted that $a$ is positive. Then $y^2>a^2$, i.e. $y$ has a larger magnitude if $y$ is positive and $y>a$ or $y$ is negative, i.e. $|y|=-y>a$, which is also $y<-a$.