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Category: Algebra

a-b-c-are-positive-real-numbers-And-1-a-b-1-1-b-c-1-1-a-c-1-1-prove-that-a-b-c-ab-bc-ac-

Question Number 193965 by Subhi last updated on 24/Jun/23 $${a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${And} \\ $$$$\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}+\frac{\mathrm{1}}{{b}+{c}+\mathrm{1}}+\frac{\mathrm{1}}{{a}+{c}+\mathrm{1}}\geqslant\mathrm{1} \\ $$$${prove}\:{that}\:{a}+{b}+{c}\geqslant{ab}+{bc}+{ac} \\ $$ Answered by Subhi last updated on 26/Jun/23…

Question-193962

Question Number 193962 by Abdullahrussell last updated on 24/Jun/23 Answered by AST last updated on 24/Jun/23 $$\Rightarrow{f}\left({x}\right)={Q}\left({x}\right)\left({x}−{a}\right)\left({x}−{b}\right)+{R}\Rightarrow{f}\left({a}\right)={R},{f}\left({b}\right)={R} \\ $$$$\Rightarrow\frac{{f}\left({a}\right)\left({x}−{b}\right)−{f}\left({b}\right)\left({x}−{a}\right)}{{a}−{b}}=\frac{{R}\left({x}−{b}−{x}+{a}\right)}{{a}−{b}}={R} \\ $$ Answered by cortano12 last…

Question-193886

Question Number 193886 by Rupesh123 last updated on 22/Jun/23 Answered by MM42 last updated on 22/Jun/23 $${a}=\mathrm{521}\left(\mathrm{521}^{{n}} −\mathrm{521}^{{n}−\mathrm{1}} +\mathrm{1}\right)=\mathrm{521}{m} \\ $$$$“\mathrm{521}''\:{is}\:{prime}\:{number}.{therefore}\:“{m}'' \\ $$$$\:{must}\:{be}\:{a}\:{multiple}\:“\mathrm{521}''. \\ $$$${which}\:{is}\:{valid}\:{only}\:{for}\:\:“{n}=\mathrm{1}''…

a-b-c-d-e-f-are-real-numbers-prove-a-b-c-b-c-d-c-d-e-d-e-f-e-f-a-f-a-b-3-

Question Number 193875 by Subhi last updated on 22/Jun/23 $${a},{b},{c},{d},{e},{f},\:{are}\:+\:{real}\:{numbers} \\ $$$${prove}: \\ $$$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{d}}+\frac{{c}}{{d}+{e}}+\frac{{d}}{{e}+{f}}+\frac{{e}}{{f}+{a}}+\frac{{f}}{{a}+{b}}\geqslant\mathrm{3} \\ $$ Answered by AST last updated on 23/Jun/23 $$=\underset{{cyc}} {\sum}\frac{{a}^{\mathrm{2}}…

Question-193848

Question Number 193848 by Mingma last updated on 21/Jun/23 Answered by MM42 last updated on 21/Jun/23 $$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}×…×\:?\:×\frac{\mathrm{2024}}{\mathrm{2025}}\: \\ $$ Answered by witcher3 last updated on…

Let-G-be-a-finite-group-f-be-an-automorphism-of-G-such-that-f-x-x-x-e-Then-prove-that-i-g-G-x-G-such-that-g-x-1-f-x-ii-If-x-G-f-f-x-x-G-is-Abelian-

Question Number 193796 by Rajpurohith last updated on 20/Jun/23 $${Let}\:{G}\:{be}\:{a}\:{finite}\:{group},{f}\:{be}\:{an}\:{automorphism}\:{of}\:{G} \\ $$$${such}\:{that}\:{f}\left({x}\right)={x}\:\Rightarrow{x}={e}\:. \\ $$$${Then}\:{prove}\:{that}, \\ $$$$\left(\boldsymbol{{i}}\right)\forall{g}\in{G},\:\exists{x}\in{G}\:{such}\:{that}\:{g}={x}^{−\mathrm{1}} {f}\left({x}\right). \\ $$$$\left(\boldsymbol{{ii}}\right){If}\:\forall{x}\in{G}\:,\:{f}\left({f}\left({x}\right)\right)={x}\:\Rightarrow\:{G}\:{is}\:{Abelian}. \\ $$$$ \\ $$ Terms of…

Let-H-be-a-subgroup-of-R-such-that-H-1-1-contains-a-non-zero-element-Prove-that-H-is-cyclic-

Question Number 193794 by Rajpurohith last updated on 20/Jun/23 $${Let}\:{H}\:{be}\:{a}\:{subgroup}\:{of}\:\left(\mathbb{R},+\right)\:{such}\:{that}\:{H}\cap\left[−\mathrm{1},\mathrm{1}\right]\: \\ $$$${contains}\:{a}\:{non}\:{zero}\:{element}. \\ $$$${Prove}\:{that}\:{H}\:{is}\:{cyclic}. \\ $$ Answered by TheHoneyCat last updated on 28/Jun/23 $$ \\…