Let us prove that \(\sqrt 4 \) is irrational:

Let us assume that \(\sqrt 4\) is rational, then it can be expressed in the form of \(\frac pq\) where \(p,q\) have no factors in common, thus \(\frac pq = \sqrt 4\). or \(\color{red}{p^2 /q^2 = 4} \) then \(p^2 = 4q^2 \), \(\color{blue} {\text{with } p \text{ has a factor of 4}} \). Let \(p = 4m \), then \(\color{green}{(4m)^2 = 4q^2 } \rightarrow 16m^2 = 4q^2, 4m^2 = q^2 \), then \( \color{orange}{q \text{ also has a factor of 4}} \). But this contradicts the fact that \(p,q\) are coprimes. Thus the assumption that \(\sqrt 4 \) is rational is wrong. Hence, \(\sqrt 4 \) is also irrational.

This concludes that \(\sqrt 4 = 2 \) is an irrational number.

Which of these above colored equations are not necessarily correct?

"1" represents \(\color{red}{red}\), "2" represents \(\color{blue}{blue}\), "3" represents \(\color{green}{green}\), "4" represents \(\color{orange}{orange}\).

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