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This example is from Discrete Math and its Applications enter image description here

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I understand the steps the author is taking. First he assumes sqrt(2) is rational meaning that there exists integers a, and b such that sqrt(2) = a/b, b!= 0, and that a and b have no common factors other than 1(lowest term). He takes algebraic steps to show that if sqrt(2) = a/b, a will be even and b will be even, meaning that this is a contradiction of the lowest term idea. By proof by contradiction, therefore the author has shown that sqrt(2) is irrational, or rational.

I am trying to apply the the same proof to show that sqrt(20) is irrational. Here is my work so far enter image description here

I took the same steps as the author but I arrived at 10b^2 = 2c^2 while the author arrived at b^2 = 2c^2. The author was able to conclude that b must be even because the square of an odd must be an odd and the square of an even. I am not so sure I can conclude the same thing. What I can say is that 10b^2 will be even but b could be odd or even because even(10) * ood and even(10) * even is both even. What step should I take instead to show that sqrt(20) is irrational with a proof by contradiction?

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    $\begingroup$ hint: if $\sqrt{20}$ is irrational, then $\sqrt{5}$ is irrational. $\endgroup$
    – daPollak
    Commented Jan 24, 2015 at 21:59
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    $\begingroup$ HINT: $\sqrt{20}=\sqrt{4*5}=2\sqrt{5}$. So you need to prove that $\sqrt{5}$ is irrational. $\endgroup$
    – Mufasa
    Commented Jan 24, 2015 at 22:00
  • $\begingroup$ I don't get how showing sqrt(5) is irrational would show sqrt(20) is irrational as well... $\endgroup$ Commented Jan 24, 2015 at 22:12
  • $\begingroup$ because $\sqrt{20}$ is equal to a rational number (i.e. 2) times $\sqrt{5}$. So, if $\sqrt{5}$ can be shown to be irrational then any rational multiple of this is also irrational. See here for some properties of irrational numbers. $\endgroup$
    – Mufasa
    Commented Jan 24, 2015 at 22:16

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The matter of "evenness" (which means nothing more than being a multiple of $2$) is something that was important in the original problem regarding $\sqrt 2$. In your case, you should not necessarily be concerned with evenness. Continuing from where you left off, i.e., $10b^2 = 2c^2$, we can divide both sides by $2$ to get: $$5b^2 = c^2.$$ From here you can see that the important number is not $2$, but $5$. Following the same reasoning as the example, it must be the case that $c$ is a multiple of $5$, so $c = 5d$ for some $d$. We then have: $$5b^2 = 25d^2.$$ Consequently, $$b^2 = 5d^2$$ and we can continue in a similar fashion to arrive at a contradiction.

Now, having realized that we're not really concerned with $a$ being even or odd, let's simplify the proof by going back to the line $20b^2=a^2$. Factoring, this is $4\cdot5b^2 = a^2$, from which we conclude that $a$ is a multiple of $5$, and the proof continues as above.

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