prove that \sqrt2 is irrational
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Answer:
A proof that the square root of 2 is irrational. Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction.
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Answer:
Step-by-step explanation:
Lets assume [tex]\sqrt{2}[/tex] is rational
Hence [tex]\sqrt{2}[/tex]= a/b where a and b are two co prime numbers
[tex]\sqrt{2}[/tex]b=a
2[tex]b^{2}[/tex]=[tex]a^{2}[/tex]
2 | [tex]a^{2}[/tex]
2 | a hence 2 divides a
hence a=2c where c is any constant
[tex]\sqrt{2}[/tex]b=2c
2[tex]b^{2}[/tex]=4[tex]c^{2}[/tex]
[tex]b^{2}[/tex]=2[tex]c^{2}[/tex]
2 | [tex]b^{2}[/tex]
2 | b
hence 2 divides b
but a and b are two co primes
hence our assumption was wrong
[tex]\sqrt{2}[/tex] is irrational