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Given the equation

$\sqrt{11} +\sqrt{44}= \sqrt{99}$

Why is squaring each individual term not allowed? Doing so we get 11+44=99 which is incorrect. Is it because $\sqrt{11} +\sqrt{44}$ is considered a term grouped by addition and therefore to be treated as a single entity instead of two parts?

Sphygmomanometer
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5 Answers5

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It is because 'squaring' is not linear.

That means that the function $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^2$ has not the property that $f(a+b)=f(a)+f(b)$ since $f(a+b)=(a+b)^2=a^2+2ab+b^2\neq a^2+b^2=f(a)+f(b)$.

You can square both sides, but you have to be more specific.

If you square (what can be done since both sides are positive) we get:

$(\sqrt{11}+\sqrt{44})^2=(\sqrt{99})^2$

Which leads to:

$11+44+2\sqrt{11}\sqrt{44}=99$

So $2\sqrt{11\cdot 44}=44$

$\sqrt{484}=22\checkmark$

Cornman
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  • I think the notation in this answer is very likely to be over the OP's level of knowledge – MathTrain Sep 04 '19 at 05:22
  • @MathTrain Could be. I view my answer as seperated into two parts. The first half giving some insight in operations that have the confused property and the second half gives the correct approach how to deal with it. So I think more then just the OP could learn something from it. An answer does not have to be 100% suited for an OP. There are more people reading such answers and this question already has 5, so I think the OP is pleased. – Cornman Sep 04 '19 at 05:25
  • @MathTrain Yeah a little bit, but its ok since I am used google to look up the function notation. Although embarrassingly enough I'm in Calc 3 and making these kinds of algebra mistakes. Bit of a letdown since I am interested in math, guess I got to really hit the books. – Sphygmomanometer Sep 04 '19 at 05:27
  • @Sphygmomanometer Don't worry about it, that's the point of learning. I think taking a class on, or doing research in, some of the fundamentals might help you, such as an introductory course in proofs ( I could point you to some resources if you want). It doesn't mean you're not good at it, just that you haven't gone all the way through learning the concepts yet. The fact that you understood the notation after a quick google is honestly quite a good sign – MathTrain Sep 04 '19 at 05:32
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    @Sphygmomanometer As long as you are fixing your gaps, it is fine. Never give up on math, and you will become better and better. I think most people have done this mistake at least once in their live. I have to admit that it should be uncommen among math students, but it is no big deal. Keep in mind that this identity is correct in a field of characteristic 2, like $\mathbb{F}_2$. Because here is $2=0$, so $(a+b)^2=a^2+b^2$. :) Maybe you remembered that, so called, Frobeinus homomorphism. – Cornman Sep 04 '19 at 05:32
  • @Cornman Very good point, I do forget that sometimes the answers are for everybody and not just the OP! But it seems he/she was able to glean much from your answer either way :) – MathTrain Sep 04 '19 at 05:33
  • Thanks for the encouragement guys and yes @MathTrain I would appreciate pointers. – Sphygmomanometer Sep 04 '19 at 05:53
  • @Sphygmomanometer I'll update with some tomorrow when I get a chance – MathTrain Sep 04 '19 at 06:03
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The simplest reason that it isn't allowed is that it simply doesn't work. If you take the numbers $\sqrt{11}$ and $\sqrt{44}$, then add them, and then square the result, you do not get $11+99$, you get $11+44+2\sqrt{11\cdot 44}=11+44+2\cdot 22=99$.

We can get the correct result as shown above by using a different method. It is the distributive property, where we take each piece and multiply it out. To do the above calculation more carefully:

$$(\sqrt{11}+\sqrt{44})^2=(\sqrt{11}+\sqrt{44})\cdot (\sqrt{11}+\sqrt{44})$$$$= (\sqrt{11}+\sqrt{44})\cdot (\sqrt{11})+(\sqrt{11}+\sqrt{44})\cdot (\sqrt{44})$$ $$=11+44+2\sqrt{11\cdot 44}=11+44+(2\cdot 22)=99$$

This works for any numbers at all, not just 11 and 44. For instance:

$$(\sqrt{4}+\sqrt{49})^2=(\sqrt{4}+\sqrt{49})\cdot (\sqrt{4}+\sqrt{49})$$$$= (\sqrt{4}+\sqrt{49})\cdot (\sqrt{4})+(\sqrt{4}+\sqrt{49})\cdot (\sqrt{49})$$ $$=4+49+2\sqrt{4\cdot 49}=4+49+(2\cdot 14)=81$$

Which you may note agrees with the more obvious: $(2+7)^2=81$. If we assumed that the square can "distribute" as in your conjecture, we would get the incorrect statement $(2+7)^2=4+49=53$.

So even though it looks as though it may be reasonable for the square to distribute, it cannot do that, or else we would have nonsense.

This should, I hope, make it a bit more clear that a) the distributive property (for multiplying) is very useful and cool, and a very interesting property for multiplication to have AND b) the rules for squaring and other powers are actually direct consequences of the distributive property. Squaring means multiplying a number by itself, and the distributive property determines what that must be.

MathTrain
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In an equation, you are allowed to do the same operation to both sides. So squaring the whole left side gives you $(\sqrt{11}+\sqrt{44})^2$. And "unfortunately", this is not the same as $\sqrt{11}^2+\sqrt{44}^2$. So yes, in a sense, you have to think of the entire left-hand-side as a single term in this respect.

Arthur
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Because $(a+b)^2=a^2+2ab+b^2$. There’s a cross term $2ab$.

Alex R.
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You can think it Geometrically. Assume you cut three pieces of paper in square shape of 2 cm , 3 cm and 5 cm. According to you Both small paper should fully cover bigger one. but if you Observe , there is some area left which can easily be seen is $2(2.3)$[two times a rectangle]

vrintle
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Rishi
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