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Suppose we have to work with approximate numbers up to the second order, i.e., with two decimal places. Now consider $a=5/3=1.\bar6\approx1.67$.

Now we have to solve $$b=1+a\tag1$$ so we have to insert $a\approx1.67$ in that equation.

However, I have doubts about how an approximately could be written:

$$b\boldsymbol{\color{red}=}1+1.67\implies b\approx2.67\tag2$$

$$b\boldsymbol{\color{red}\approx}1+1.67\implies b\approx2.67.\tag3$$

Note that I have doubts when the value is already entered, not in $(1)$.

What would you do?

Thanks!

manooooh
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  • Perhaps the correct option (and the one I am currently using) is $(3)$ because of the transitivity of the symbols of equality $=$ and approximately $\approx$. – manooooh Sep 24 '18 at 22:44
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    Because you solve $b=1+a$ the exact number is not $1+1.67$, so $(3)$ would be the correct. You have $b=1+a\approx 1+1.67=2.67\implies b\approx 2.67$, $=$ "carries" the properties of $\approx$. Note that in math $=$ means (usually) that the left side and the right side are exactly the same object – ℋolo Sep 24 '18 at 22:46
  • @Holo that makes sense! – manooooh Sep 24 '18 at 22:47

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