how to prove that for every non negative n , sqrt(n) is either an integer or an irrational number .
The Higgs boson is an elementary particle in the Standard Model of particle physics, often referred to as the "God particle." It plays a crucial role in explaining why other particles have mass. Here's a breakdown of its significance: 1. The Higgs Field: The Higgs boson is associated with the HiggsRead more
The Higgs boson is an elementary particle in the Standard Model of particle physics, often referred to as the “God particle.” It plays a crucial role in explaining why other particles have mass. Here’s a breakdown of its significance:
1. The Higgs Field: The Higgs boson is associated with the Higgs field, an invisible energy field that permeates the entire universe. According to the Standard Model, this field is responsible for giving mass to elementary particles.
2. Mass Acquisition: When particles interact with the Higgs field, they acquire mass. The more strongly a particle interacts with the field, the more massive it becomes. Particles that do not interact with the Higgs field, like photons, remain massless.
3. Higgs Boson as Evidence: The Higgs boson is the quantum excitation of the Higgs field, meaning it is the particle form of the field. Its discovery provided direct evidence that the Higgs field exists and operates as theorized.
4. Discovery: The Higgs boson was discovered in 2012 by scientists at the Large Hadron Collider (LHC) at CERN. This discovery confirmed the mechanism that explains how particles acquire mass, a cornerstone of the Standard Model.
5. Nobel Prize: The discovery of the Higgs boson led to the awarding of the Nobel Prize in Physics in 2013 to François Englert and Peter Higgs, who had proposed the existence of the Higgs mechanism independently in the 1960s.
The Higgs boson is a fundamental particle that confirms the mechanism by which particles acquire mass, thus playing a critical role in our understanding of the universe’s fundamental structure.
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For every non-negative integer \( n \), the square root \( \sqrt{n} \) is either an integer or an irrational number. Proof (by contradiction): Suppose \( \sqrt{n} \) is rational but **not** an integer. Then there exist integers \( a \) and \( b \), with \( \gcd(a, b) = 1 \) and \( b \ne 1 \), such tRead more
For every non-negative integer \( n \), the square root \( \sqrt{n} \) is either an integer or an irrational number.
Proof (by contradiction):
Suppose \( \sqrt{n} \) is rational but **not** an integer.
Then there exist integers \( a \) and \( b \), with \( \gcd(a, b) = 1 \) and \( b \ne 1 \), such that:
\[
\sqrt{n} = \frac{a}{b}
\]
Squaring both sides:
\[
n = \left( \frac{a}{b} \right)^2 = \frac{a^2}{b^2}
\Rightarrow a^2 = n b^2
\]
This implies that \( b^2 \) divides \( a^2 \). But since \( \gcd(a, b) = 1 \), it follows that \( \gcd(a^2, b^2) = 1 \) as well. Hence, the only way \( b^2 \mid a^2 \) can be true is if \( b^2 = 1 \), which means:
\[
b = 1
\Rightarrow \sqrt{n} = \frac{a}{1} = a \in \mathbb{Z}
\]
This contradicts our assumption that \( \sqrt{n} \) is rational **but not** an integer.
Conclusion:
If \( \sqrt{n} \) is rational, then it must be an integer.
Therefore, if \( \sqrt{n} \) is not an integer, it must be irrational.
\[
See less\boxed{\text{For all } n \in \mathbb{N}_0,\ \sqrt{n} \in \mathbb{Z} \cup (\mathbb{R} \setminus \mathbb{Q})}
\]