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Question Number 56243 by Kunal12588 last updated on 12/Mar/19

6/2×5 which one correct 6/2×5 6/2×5 =3×5 =6/10 =15 =0.6

$$\mathrm{6}/\mathrm{2}×\mathrm{5} \\ $$$$\:{which}\:{one}\:{correct} \\ $$$$\mathrm{6}/\mathrm{2}×\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}/\mathrm{2}×\mathrm{5} \\ $$$$=\mathrm{3}×\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6}/\mathrm{10} \\ $$$$=\mathrm{15}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{6} \\ $$

Answered by MJS last updated on 12/Mar/19

one after the other. for your 2^(nd) version you need brackets 6/2×5=3×5=15 6/(2×5)=6/10=0.6 it′s similar to this: 6−2+5=4+5=9 6−(2+5)=6−7=−1

$$\mathrm{one}\:\mathrm{after}\:\mathrm{the}\:\mathrm{other}.\:\mathrm{for}\:\mathrm{your}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{version}\:\mathrm{you} \\ $$$$\mathrm{need}\:\mathrm{brackets} \\ $$$$\mathrm{6}/\mathrm{2}×\mathrm{5}=\mathrm{3}×\mathrm{5}=\mathrm{15} \\ $$$$\mathrm{6}/\left(\mathrm{2}×\mathrm{5}\right)=\mathrm{6}/\mathrm{10}=\mathrm{0}.\mathrm{6} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{similar}\:\mathrm{to}\:\mathrm{this}: \\ $$$$\mathrm{6}−\mathrm{2}+\mathrm{5}=\mathrm{4}+\mathrm{5}=\mathrm{9} \\ $$$$\mathrm{6}−\left(\mathrm{2}+\mathrm{5}\right)=\mathrm{6}−\mathrm{7}=−\mathrm{1} \\ $$

Commented by Kunal12588 last updated on 12/Mar/19

thanks sir.

$${thanks}\:{sir}. \\ $$

Commented by Kunal12588 last updated on 12/Mar/19

sir we can change 6−2+5=6−(2−5) is there any way for 6/2×5 for multiplying first then divide.

$${sir}\:{we}\:{can}\:{change} \\ $$$$\mathrm{6}−\mathrm{2}+\mathrm{5}=\mathrm{6}−\left(\mathrm{2}−\mathrm{5}\right) \\ $$$${is}\:{there}\:{any}\:{way}\:{for}\:\mathrm{6}/\mathrm{2}×\mathrm{5}\:{for}\:{multiplying} \\ $$$${first}\:{then}\:{divide}.\: \\ $$

Commented by MJS last updated on 12/Mar/19

it′s similar again: 6/2×5=6/(2/5) but it′s also commutative: a−b is originally defined as addition of the inverse element: a+(−b) similar a/b=a×b^(−1) 6+(−2)+5=6+5+(−2)=(−2)+5+6=... 6×2^(−1) ×5=6×5×2^(−1) =2^(−1) ×5×6=...

$$\mathrm{it}'\mathrm{s}\:\mathrm{similar}\:\mathrm{again}: \\ $$$$\mathrm{6}/\mathrm{2}×\mathrm{5}=\mathrm{6}/\left(\mathrm{2}/\mathrm{5}\right) \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{commutative}: \\ $$$${a}−{b}\:\mathrm{is}\:\mathrm{originally}\:\mathrm{defined}\:\mathrm{as}\:\mathrm{addition}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inverse}\:\mathrm{element}:\:{a}+\left(−{b}\right) \\ $$$$\mathrm{similar} \\ $$$${a}/{b}={a}×{b}^{−\mathrm{1}} \\ $$$$\mathrm{6}+\left(−\mathrm{2}\right)+\mathrm{5}=\mathrm{6}+\mathrm{5}+\left(−\mathrm{2}\right)=\left(−\mathrm{2}\right)+\mathrm{5}+\mathrm{6}=... \\ $$$$\mathrm{6}×\mathrm{2}^{−\mathrm{1}} ×\mathrm{5}=\mathrm{6}×\mathrm{5}×\mathrm{2}^{−\mathrm{1}} =\mathrm{2}^{−\mathrm{1}} ×\mathrm{5}×\mathrm{6}=... \\ $$

Commented by Kunal12588 last updated on 13/Mar/19

great thanks once again.

$${great}\:{thanks}\:{once}\:{again}. \\ $$

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