Is my proof for the sum of two square roots being irrational correct?

Asked Oct 15 '15 at 08:22

Active Oct 15 '15 at 15:18

Viewed 151 times

Prove that the following number is irrational:

$$\sqrt { 5 } +\sqrt { 3 } $$

Steps I took:

Proof by contradiction: let us assume that $\sqrt { 5 } +\sqrt { 3 } $ is rational.

If $\sqrt { 5 } +\sqrt { 3 } $ is rational, so is $(\sqrt { 5 } +\sqrt { 3 } )^{ 2 }$

$$(\sqrt { 5 } +\sqrt { 3 } )^{ 2 }=5+2\sqrt { 15 } +3= 2\sqrt { 15 } +8$$

However, this is a contradiction since $\sqrt { 15 } $ is irrational.

Is my proof good? What needs to be added and how can it be made more complete if it already isn't. Please note that I already proved that $\sqrt { 15 } $ is irrational in part (a) of the problem. This is part(b). That is why I didn't bother to prove it again. I guess you could say the irrationality of $\sqrt { 15 } $ was given when I started the proof.

edited Oct 15 '15 at 15:18

asked Oct 15 '15 at 08:22

Cherry_Developer

4,239

6

your proof seems good. – GAVD Oct 15 '15 at 08:23
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A little mistake. You should replace the sign $\Rightarrow$ by $=$. – user37238 Oct 15 '15 at 08:31
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Your proof is correct but the method does not always work. What if you had to prove that $\sqrt{8}+\sqrt{2}$ is irrational? The square is $8+8+2=18$. – uniquesolution Oct 15 '15 at 08:37
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@user37238 better? – Cherry_Developer Oct 15 '15 at 09:26
@uniquesolution what would you recommend? – Cherry_Developer Oct 15 '15 at 09:26
@uniquesolution in that case his approach would work again: $\sqrt 8+\sqrt 2=3\sqrt 2$. – Math-fun Oct 15 '15 at 09:37
@Math-fun -- his approach was to square the number, not what you wrote. – uniquesolution Oct 15 '15 at 09:40

Is my proof for the sum of two square roots being irrational correct?

0 Answers0