Question

1. is $sqrt{2}$ an irrational number?

click on a or b in the box to select your answer
a) yes
b) no
1 pt

Explanation:

Step1: Define rational numbers

A rational number can be written as $\frac{p}{q}$, where $p,q$ are integers, $q
eq0$, and $\gcd(p,q)=1$.

Step2: Assume $\sqrt{2}$ is rational

Suppose $\sqrt{2}=\frac{p}{q}$. Square both sides: $2=\frac{p^2}{q^2} \implies p^2=2q^2$.

Step3: Analyze parity of $p$

$p^2$ is even, so $p$ is even. Let $p=2k$, $k\in\mathbb{Z}$.

Step4: Substitute $p=2k$ into equation

$(2k)^2=2q^2 \implies 4k^2=2q^2 \implies q^2=2k^2$. Thus $q$ is even.

Step5: Contradiction conclusion

$p$ and $q$ are both even, so $\gcd(p,q)\geq2$, contradicting $\gcd(p,q)=1$. So $\sqrt{2}$ is irrational.

Answer:

A) Yes