given a^2 <1 now a<(√1) or a<±1 ∴ a<1 and a<−1 but but its false we know if a^2 <1 so −1<a<1 so my question is why this is happening at all.


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Question Number 37738 by kunal1234523 last updated on 17/Jun/18

$given a^{2} < 1$ $now$ $a < \sqrt{1}$ $or a < \pm 1$ $∴ a < 1 and a < - 1 but but its false we know$ $i f a^{2} < 1 so - 1 < a < 1$ $s o m y q u e s t i o n i s w h y t h i s i s h a p p e n i n g a t a l l .$
Commented byprakash jain last updated on 17/Jun/18

$\sqrt{a^{2}} =∣ a ∣$ $\sqrt{a^{2}} \neq a$
Commented byMrW3 last updated on 17/Jun/18

$f r o m a^{2} < 1 y o u c a n g e t$ $a < \sqrt{1} (t o b e a x a c t, i t s h o u l d b e ∣ a ∣< \sqrt{1})$ $b u t y o u c a n n o t g e t$ $a < \pm 1! (f r o m ∣ a ∣< 1, y o u s h o u l d g e t - 1 < a < 1)$
Commented bykunal1234523 last updated on 17/Jun/18

$ohh! thank you actually i had not started learning$ $about absolute value$
Commented byRasheed.Sindhi last updated on 17/Jun/18

$a^{2} < 1 \Rightarrow a^{2} - 1 < 0 \Rightarrow (a - 1) (a + 1) < 0$ $\Rightarrow (a - 1 < 0 \land a + 1 > 0) ∣ (a - 1 > 0 \land a + 1 < 0)$ $\Rightarrow (a < 1 \land a > - 1) ∣ (a > 1 \land a < - 1)$ $\Rightarrow - 1 < a < 1 ∣ (a > 1 \land a < - 1) (Impossible)$ $\Rightarrow - 1 < a < 1$
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