Can I prove that both 2^1/2 and 2^1/3 are irrational, so the sum of them is also irrational?
Thanks~~
Thanks~~
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Proving that both numbers are the irrational is not necessarily the same as showing that their sum is irrational. For example, if we took the numbers π and -π, then each is irrational but the sum is 0, which is rational.
The rules are:
rational+rational = rational
rational + irrational = irrational
irrational + irrational = ?
A similar rule applies to multiplication, except when 0 is involved.
Presuming that you could show that √2 is irrational, here are the extra steps you would need:
Proof by contradiction –
Suppose 2^1/2 + 2^1/3 = q is rational
Then we can say
2^(1/3) = q - 2^(1/2) ... cube both sides
2 = (q - √2)^3 = q^3 - 3q^2√2 + 3q*2 - 2√2... simplify
2 = q^3 + 6q - 3q^2√2 - 2√2
2 = q(q^2 + 6) - √2(3q^2 + 2)... now solve for √2
√2(3q^2 + 2) + 2 = q(q^2 + 6)
√2(3q^2 + 2)= q(q^2 + 6) - 2
√2 = (q(q^2 + 6) - 2)/(3q^2 + 2)
If q is rational, then the object on the right must be rational, thus √2 must be rational. However, we know √2 to be irrational, so we have a contradiction
So, q is irrational.
Note: the reason that last bit has to be rational is that adding and multiplying rational numbers with other rational numbers will give you another rational number, no matter how many times you do it or in what order.
Hope that helps!
The rules are:
rational+rational = rational
rational + irrational = irrational
irrational + irrational = ?
A similar rule applies to multiplication, except when 0 is involved.
Presuming that you could show that √2 is irrational, here are the extra steps you would need:
Proof by contradiction –
Suppose 2^1/2 + 2^1/3 = q is rational
Then we can say
2^(1/3) = q - 2^(1/2) ... cube both sides
2 = (q - √2)^3 = q^3 - 3q^2√2 + 3q*2 - 2√2... simplify
2 = q^3 + 6q - 3q^2√2 - 2√2
2 = q(q^2 + 6) - √2(3q^2 + 2)... now solve for √2
√2(3q^2 + 2) + 2 = q(q^2 + 6)
√2(3q^2 + 2)= q(q^2 + 6) - 2
√2 = (q(q^2 + 6) - 2)/(3q^2 + 2)
If q is rational, then the object on the right must be rational, thus √2 must be rational. However, we know √2 to be irrational, so we have a contradiction
So, q is irrational.
Note: the reason that last bit has to be rational is that adding and multiplying rational numbers with other rational numbers will give you another rational number, no matter how many times you do it or in what order.
Hope that helps!
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Hey Ben,
That question has a hint: Look at the numbers of the form p/q√2.
That question has a hint: Look at the numbers of the form p/q√2.
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I think you mean to say that a and b are rational numbers. Anyway, I'm limited here so I'll send you my answer as a message. Check your email!
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...Or the question that you posted anyway.
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a+ (b-a)/sqrt(2) is irrational
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