# Thread: Sqrt of 4 is irrational

1. @Mukilab: Your most recent post highlights the difference between "X is cheating" and "X feels like cheating." Maybe what I should have said is that my proof shouldn't feel like cheating. When there is a simple proof available, why use a more complex one?

2. I agree with undefined. If you wanted a clever proof (ie, one that doesn't use the fact that the asquare root of 4 is known to be 2), you need to say so.

3. good point, thank you for the information.

Would either be marked higher in an exam?

4. undefined @ 13: Well, some people might have a problem with $5=\pm 5$, if they're using equality as being the same thing as identity. 5 is not identically equal to plus or minus 5. In effect, what I'm really doing here is a short-hand notation for $(5=5)\vee(5=-5)$, where this equality does mean identity.

5. Originally Posted by Ackbeet
undefined @ 13: Well, some people might have a problem with $5=\pm 5$, if they're using equality as being the same thing as identity. 5 is not identically equal to plus or minus 5. In effect, what I'm really doing here is a short-hand notation for $(5=5)\vee(5=-5)$, where this equality does mean identity.
I think $5=\pm 5$ universally stands for $(5=5)\vee(5=-5)$. Reference (Wikipedia - Plus-minus sign - Subheading: Shorthand).

Originally Posted by Mukilab
good point, thank you for the information.

Would either be marked higher in an exam?
As long as you present a valid proof, you should get full marks. I think it would be unreasonable for a teacher/professor/grader to grade otherwise. The type of proof should only matter if the problem statement specifies a particular type of proof.

6. Fair point.

7. $\sqrt{4} = -2$ is false.

for all $x \in \mathbb{R}$ , $\sqrt{x^2} = |x| \ge 0$

8. What if I'm on a different branch of the square root function? You're correct, but that doesn't alter the fact that $\sqrt{4}=\pm 2$ is a true statement. If you think about the logical OR connective, with a boolean variable $x$, then it is the case that $x\vee\text{false}=x$.

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# show root 4is irrational

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