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Question Number 105207 by Anindita last updated on 26/Jul/20

99−98((98)/(99)) = ? Can you solve this?

$$\mathrm{99}−\mathrm{98}\frac{\mathrm{98}}{\mathrm{99}}\:=\:? \\ $$$$\boldsymbol{{Can}}\:\boldsymbol{{you}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}? \\ $$

Commented by qwerty111 last updated on 26/Jul/20

98+((99)/(99))−98−((98)/(99)) = (1/(99))

$$\mathrm{98}+\frac{\mathrm{99}}{\mathrm{99}}−\mathrm{98}−\frac{\mathrm{98}}{\mathrm{99}}\:=\:\frac{\mathrm{1}}{\mathrm{99}} \\ $$

Commented by JDamian last updated on 26/Jul/20

do you assume a(b/c) means a+(b/c)?

$${do}\:{you}\:{assume}\:\:\:{a}\frac{{b}}{{c}}\:\:{means}\:\:\:{a}+\frac{{b}}{{c}}? \\ $$

Commented by JDamian last updated on 27/Jul/20

I never used/understood a(b/c) = a+(b/c) in algebra/maths texts. I only found in, for instance, cooking recipes books (“2 1/2 teaspoon”)

$${I}\:{never}\:{used}/{understood}\:\:{a}\frac{{b}}{{c}}\:=\:{a}+\frac{{b}}{{c}}\:{in} \\ $$$${algebra}/{maths}\:{texts}.\:{I}\:{only}\:{found}\:{in},\:{for} \\ $$$${instance},\:{cooking}\:{recipes}\:{books}\:\left(``\mathrm{2}\:\mathrm{1}/\mathrm{2}\right. \\ $$$$\left.{teaspoon}''\right) \\ $$

Answered by Aziztisffola last updated on 26/Jul/20

99((99)/(99))−98((98)/(99))=99((98+1)/(99))−98((98)/(99)) =(98+1)((98+1)/(99))−98((98)/(99))=98((98)/(99))+((2×98)/(99))+(1/(99))−98((98)/(99)) =((2×98+1)/(99))=((196)/(99))

$$\mathrm{99}\frac{\mathrm{99}}{\mathrm{99}}−\mathrm{98}\frac{\mathrm{98}}{\mathrm{99}}=\mathrm{99}\frac{\mathrm{98}+\mathrm{1}}{\mathrm{99}}−\mathrm{98}\frac{\mathrm{98}}{\mathrm{99}} \\ $$$$=\left(\mathrm{98}+\mathrm{1}\right)\frac{\mathrm{98}+\mathrm{1}}{\mathrm{99}}−\mathrm{98}\frac{\mathrm{98}}{\mathrm{99}}=\mathrm{98}\frac{\mathrm{98}}{\mathrm{99}}+\frac{\mathrm{2}×\mathrm{98}}{\mathrm{99}}+\frac{\mathrm{1}}{\mathrm{99}}−\mathrm{98}\frac{\mathrm{98}}{\mathrm{99}} \\ $$$$=\frac{\mathrm{2}×\mathrm{98}+\mathrm{1}}{\mathrm{99}}=\frac{\mathrm{196}}{\mathrm{99}} \\ $$

Commented by Aziztisffola last updated on 27/Jul/20

I know that a+(b/c)≠a×(b/c) ax=a×x not a+x.

$$\mathrm{I}\:\mathrm{know}\:\mathrm{that}\:\mathrm{a}+\frac{\mathrm{b}}{\mathrm{c}}\neq\mathrm{a}×\frac{\mathrm{b}}{\mathrm{c}} \\ $$$$\mathrm{ax}=\mathrm{a}×\mathrm{x}\:\mathrm{not}\:\mathrm{a}+\mathrm{x}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 27/Jul/20

98×((98)/(99))+((2×98)/(99))+(1/(99))−98((98)/(99)) ≠ ((2×98+1)/(99)) as 98×((98)/(99)) ≠98((98)/(99))

$$\mathrm{98}×\frac{\mathrm{98}}{\mathrm{99}}+\frac{\mathrm{2}×\mathrm{98}}{\mathrm{99}}+\frac{\mathrm{1}}{\mathrm{99}}−\mathrm{98}\frac{\mathrm{98}}{\mathrm{99}}\:\neq\:\frac{\mathrm{2}×\mathrm{98}+\mathrm{1}}{\mathrm{99}} \\ $$$$\mathrm{as}\:\mathrm{98}×\frac{\mathrm{98}}{\mathrm{99}}\:\neq\mathrm{98}\frac{\mathrm{98}}{\mathrm{99}} \\ $$

Commented by PRITHWISH SEN 2 last updated on 27/Jul/20

2(1/3)=(7/3) 2(1/3) = 2+(1/3)=(7/3) 2(1/3)=2×(1/3)= (2/3) ∴ 2(1/3)= 2+(1/3) ≠ 2×(1/3) ∴ you are wrong

$$\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}\:=\:\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\therefore\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}=\:\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}}\:\neq\:\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\therefore\:\mathrm{you}\:\mathrm{are}\:\mathrm{wrong} \\ $$

Commented by qwerty111 last updated on 26/Jul/20

★ 98 ((98+1)/(99 )) ★

$$\:\bigstar\:\:\:\mathrm{98}\:\frac{\mathrm{98}+\mathrm{1}}{\mathrm{99}\:}\:\:\:\bigstar \\ $$

Commented by Aziztisffola last updated on 26/Jul/20

Just a simple calculus.

$$\mathrm{Just}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{calculus}. \\ $$

Commented by qwerty111 last updated on 26/Jul/20

Ok

$$\boldsymbol{{Ok}} \\ $$

Commented by Aziztisffola last updated on 26/Jul/20

I assume a(b/c)=a×(b/c) not a+(b/c)

$$\mathrm{I}\:\mathrm{assume}\:\mathrm{a}\frac{\mathrm{b}}{\mathrm{c}}=\mathrm{a}×\frac{\mathrm{b}}{\mathrm{c}}\:\mathrm{not}\:\mathrm{a}+\frac{\mathrm{b}}{\mathrm{c}} \\ $$

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