given-a-2-lt-1-now-a-lt-1-or-a-lt-1-a-lt-1-and-a-lt-1-but-but-its-false-we-know-if-a-2-lt-1-so-1-lt-a-lt-1-so-my-question-is-why-this-is-happening-at-all-

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 by prakash jain last updated on 17/Jun/18

\sqrt{a^{2}} =∣ a ∣

\sqrt{a^{2}} \neq a

Commented by MrW3 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 by kunal1234523 last updated on 17/Jun/18

ohh! thank you actually i had not started learning

about absolute value

Commented by Rasheed.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