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Question Number 185973 by normans last updated on 30/Jan/23

             [prove that;]                1 + 2 = 3

$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\left[\boldsymbol{{prove}}\:\boldsymbol{{that}};\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\mathrm{2}\:=\:\mathrm{3} \\ $$$$ \\ $$

Commented by MJS_new last updated on 30/Jan/23

if “1”=1∈N∧“2”=2∈N∧“3”=3∈N∧(“+”  is the common addition)∧(“=” is the common  equality sign):  lhs 1+2=3  rhs 3  lhs=rhs

$$\mathrm{if}\:``\mathrm{1}''=\mathrm{1}\in\mathbb{N}\wedge``\mathrm{2}''=\mathrm{2}\in\mathbb{N}\wedge``\mathrm{3}''=\mathrm{3}\in\mathbb{N}\wedge\left(``+''\right. \\ $$$$\left.\mathrm{is}\:\mathrm{the}\:\mathrm{common}\:\mathrm{addition}\right)\wedge\left(``=''\:\mathrm{is}\:\mathrm{the}\:\mathrm{common}\right. \\ $$$$\left.\mathrm{equality}\:\mathrm{sign}\right): \\ $$$$\mathrm{lhs}\:\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\mathrm{rhs}\:\mathrm{3} \\ $$$$\mathrm{lhs}=\mathrm{rhs} \\ $$

Commented by MJS_new last updated on 30/Jan/23

questions like this one only make sense when  we specify what tools are allowed for the  proof.  i.e. “prove that −5÷4 is not defined” only  makes sense when we are limited to Z

$$\mathrm{questions}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one}\:\mathrm{only}\:\mathrm{make}\:\mathrm{sense}\:\mathrm{when} \\ $$$$\mathrm{we}\:\mathrm{specify}\:\mathrm{what}\:\mathrm{tools}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{proof}. \\ $$$$\mathrm{i}.\mathrm{e}.\:``\mathrm{prove}\:\mathrm{that}\:−\mathrm{5}\boldsymbol{\div}\mathrm{4}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}''\:\mathrm{only} \\ $$$$\mathrm{makes}\:\mathrm{sense}\:\mathrm{when}\:\mathrm{we}\:\mathrm{are}\:\mathrm{limited}\:\mathrm{to}\:\mathbb{Z} \\ $$

Answered by Mathspace last updated on 30/Jan/23

but in Z/3Z     1+2=0   first you must choose the set!

$${but}\:{in}\:{Z}/\mathrm{3}{Z}\:\:\:\:\:\mathrm{1}+\mathrm{2}=\mathrm{0}\:\:\:{first}\:{you}\:{must}\:{choose}\:{the}\:{set}! \\ $$

Commented by MJS_new last updated on 30/Jan/23

yes.

$$\mathrm{yes}. \\ $$

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