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Precalculus Mathematics Homework Help
Proof that the Square Root of 2 is Irrational.
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[QUOTE="Calu, post: 4625110, member: 498278"] I am trying to prove that √2 is irrational using proof by contradiction. Here is my work so far: √2 = p/q where p & q are in their lowest terms. Where q is non-zero. 2=p[SUP]2[/SUP]/q[SUP]2[/SUP] 2q[SUP]2[/SUP] = p[SUP]2[/SUP] Which tells me that p[SUP]2[/SUP] is an even number, using the definition of an even number. We can use this definition because we know that q is an integer and an integer squared is also an integer. Hence 2q[SUP]2[/SUP] is even and as 2q[SUP][/SUP]=p[SUP]2[/SUP] we see that p[SUP]2[/SUP] is also even. Can we use the knowledge that p is an integer to say that p is also an even number? Or is there another rule that I'm not aware of? If we can use this knowledge, then we can say that p=2n, where n is some integer, using the definition of an even number. 2q[SUP]2[/SUP]=(2n)[SUP]2[/SUP] 2q[SUP]2[/SUP]=4n[SUP]2[/SUP] q[SUP]2[/SUP]=2n[SUP]2[/SUP] where n[SUP]2[/SUP] is an integer, as n was an integer from the definition of an even number. ∴ q[SUP]2[/SUP] is also an even number. If both p and q are even numbers then the fraction p/q can be simplified further and p/q is not in its lowest terms. This shows that √2 is in fact not a rational number and proves that it is irrational. [/QUOTE]
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Precalculus Mathematics Homework Help
Proof that the Square Root of 2 is Irrational.
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