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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-

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Question Number 37738 by kunal1234523 last updated on 17/Jun/18
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.
$$\mathrm{given}\:{a}^{\mathrm{2}} <\mathrm{1} \\ $$$$\mathrm{now} \\ $$$${a}<\sqrt{\mathrm{1}} \\ $$$$\mathrm{or}\:{a}<\pm\mathrm{1} \\ $$$$\therefore\:{a}<\mathrm{1}\:\mathrm{and}\:{a}<−\mathrm{1}\:\:\:\:\mathrm{but}\:\mathrm{but}\:\mathrm{its}\:\mathrm{false}\:\mathrm{we}\:\mathrm{know} \\ $$$${if}\:\mathrm{a}^{\mathrm{2}} <\mathrm{1}\:\mathrm{so}\:−\mathrm{1}<{a}<\mathrm{1}\: \\ $$$${so}\:{my}\:{question}\:{is}\:{why}\:{this}\:{is}\:{happening}\:{at}\:{all}. \\ $$
Commented by prakash jain last updated on 17/Jun/18
(√a^2 )=∣a∣ (√a^2 ) ≠a
$$\sqrt{{a}^{\mathrm{2}} }=\mid{a}\mid\: \\ $$$$\sqrt{{a}^{\mathrm{2}} }\:\neq{a} \\ $$
Commented by MrW3 last updated on 17/Jun/18
from a^2 <1 you can get a<(√1) (to be axact, it should be ∣a∣<(√1)) but you can not get a<±1 ! (from ∣a∣<1, you should get −1<a<1)
$${from}\:{a}^{\mathrm{2}} <\mathrm{1}\:{you}\:{can}\:{get} \\ $$$${a}<\sqrt{\mathrm{1}}\:\:\:\:\left({to}\:{be}\:{axact},\:{it}\:{should}\:{be}\:\mid{a}\mid<\sqrt{\mathrm{1}}\right) \\ $$$${but}\:{you}\:{can}\:{not}\:{get} \\ $$$${a}<\pm\mathrm{1}\:!\:\:\left({from}\:\mid{a}\mid<\mathrm{1},\:{you}\:{should}\:{get}\:−\mathrm{1}<{a}<\mathrm{1}\right) \\ $$
Commented by kunal1234523 last updated on 17/Jun/18
ohh! thank you actually i had not started learning about absolute value
$$\mathrm{ohh}!\:\mathrm{thank}\:\mathrm{you}\:\mathrm{actually}\:\mathrm{i}\:\mathrm{had}\:\mathrm{not}\:\mathrm{started}\:\mathrm{learning} \\ $$$$\mathrm{about}\:\mathrm{absolute}\:\mathrm{value}\: \\ $$
Commented by Rasheed.Sindhi last updated on 17/Jun/18
a^2 <1⇒a^2 −1<0⇒(a−1)(a+1)<0 ⇒(a−1<0 ∧ a+1>0) ∣ (a−1>0 ∧ a+1<0) ⇒(a<1 ∧ a>−1) ∣ (a>1 ∧ a<−1) ⇒ −1<a<1 ∣ (a>1 ∧ a<−1)(Impossible) ⇒ −1<a<1
$${a}^{\mathrm{2}} <\mathrm{1}\Rightarrow{a}^{\mathrm{2}} −\mathrm{1}<\mathrm{0}\Rightarrow\left({a}−\mathrm{1}\right)\left({a}+\mathrm{1}\right)<\mathrm{0} \\ $$$$\Rightarrow\left({a}−\mathrm{1}<\mathrm{0}\:\wedge\:{a}+\mathrm{1}>\mathrm{0}\right)\:\mid\:\left({a}−\mathrm{1}>\mathrm{0}\:\wedge\:{a}+\mathrm{1}<\mathrm{0}\right) \\ $$$$\Rightarrow\left({a}<\mathrm{1}\:\wedge\:{a}>−\mathrm{1}\right)\:\mid\:\left({a}>\mathrm{1}\:\wedge\:{a}<−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:−\mathrm{1}<{a}<\mathrm{1}\:\:\:\mid\:\:\:\left({a}>\mathrm{1}\:\wedge\:{a}<−\mathrm{1}\right)\left(\mathrm{Impossible}\right) \\ $$$$\Rightarrow\:\:−\mathrm{1}<{a}<\mathrm{1}\: \\ $$

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