Confusing proof for $\sqrt{2}$ being irrational

Question

$\sqrt{2}$ is irrational using proof by contradiction.

say $\sqrt{2}$ = $\frac{a}{b}$ where $a$ and $b$ are positive integers.

$b\sqrt{2}$ is an integer. ----[Understood]

Let $b$ denote the smallest such positive integer.----[My understanding of this is

that were are going to assume b is the smallest possible integer such that $\sqrt{2}$ = $\frac{a}{b}$, ... Understood]

Then $b^{*}$ := $b(\sqrt{2} - 1)$----[I'm not sure I understand the point that's being made here,

from creating a variable $b^{*} = a - b$ ]

Next, $b^{*}$ := $b(\sqrt{2} - 1)$ is a positive integer such that $b^{*}\sqrt{2}$ is an integer.----[ I get that ($a - b$) has to

be a positive integer, why does it follow that then $b^{*}\sqrt{2}$ is an integer?]

Lastly, $b^{*}<b$, which is a contradiction.----[I can see that given $b^{*}$ := $b(\sqrt{2} - 1)$, we then have

$b^{*}<b$, I don't get how that creates a contradiction]

Any help is appreciated, thank you.

Answer 1

So you got to $b\sqrt 2$ is an integer.

Thus $b\sqrt 2- b$ is an integer because if you subtract two integers you get an integer.

$b\sqrt 2-b = b(\sqrt 2 -1):=b^*$ is an integer.

Hopefully that clears the first issue.

Now notice that $b^*\sqrt 2 = (b\sqrt 2-b)\sqrt 2 = b\sqrt 2\sqrt 2 + b\sqrt 2 = 2b + b\sqrt 2$.

Now $b$ is an integer so $2b$ is an integer. And $b\sqrt 2$ was an integer (that was our first observation). And so integers plus integers are integers.

So $2b + b\sqrt 2= b^*\sqrt 2$ is an integer.

Hopefully that clears the second issue.

$b^*=b(\sqrt 2 - 1)$ and $0 < (\sqrt 2 - 1) < 1$. (The text should have proven this. But you claim you were okay with this...[1]) $0*b < (\sqrt 2-1)*b < 1*b$ and So $0< b^*< b$.

Why is this a contradiction?

Because you assumed $b$ was the smallest positive integer where $b\sqrt 2$ is an integer.

Consider $1,2,3,.....= \mathbb N$ and consider the list $1\sqrt 2, 2\sqrt 2, 3\sqrt 2,....$

Now either none of them are integers or there is a first item on the list that is an integer. If we assume $b\sqrt 2$ is the first on the list then $b$ is the smallest positive integer so that $b\sqrt 2=a$ is an integer.

But we have just discovered that $b^*\sqrt 2 = b(\sqrt 2-1)*\sqrt 2= 2b-b\sqrt 2=2b-a$ is a smaller integer.

That contradicts that $b$ is the smallest. Which means there can't be any smallest value where $b\sqrt 2$ is an integer.

That is the contradiction.

In my opinion, the text maybe should have defined $b$ as the smallest integer $b$ so that $b\sqrt 2$ is an integer first before defining $\sqrt 2 =\frac ab$. There's no indication that $b$ had to be the smallest positive integer. But something has to be. And if $\beta \sqrt 2 = \alpha$ is such that $\beta$ is the smallest then $\sqrt 2 =\frac{\alpha}{\beta}$.

Note.... the contradiction arises because we assume all rational numbers $r$ can be written as a fraction $\frac ab$ where $a$ is an integer and $b$ is a positive integer. This is a fair assumption as that is simply the definition of "rational number". AND WE ASSUMED there is always a smallest such integer $b$.

I think it's worth questioning our assumptions. Why do we assume all rational numbers can be expressed in such a "lowest term"?

Food for thought.

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[1] Oh what the heck. If $0\le x \le 1$ then $0 \le x^2 \le 1$ and $x^2 \ne 2$. And if $x\ge 2$ then $x^2 \ge 4$ and $x^2 \ne 2$. So if a positive $\sqrt 2$ exists at all [2] then $1 < \sqrt 2 < 2$ so $0 < \sqrt 2 -1< 1$.

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[2] Actually it's important to realize that this proof doesn't prove $\sqrt 2$ is irrational. It just proves that a rational $\sqrt 2$ does not exist. Proving that there are irrational numbers and that there is a $\sqrt 2$ is another issue altogether.

BUT that's okay. In this proof by contradiction you are assuming that $\sqrt 2 = \frac ab \in \mathbb Q$ does exist.

Answer 2

Simply calculate $$b(\sqrt2-1)\sqrt2=b(2-\sqrt2)=2b-b\sqrt2$$ which is an integer. This is actually the heart of the proof. We assumed $b$ was the smallest integer satisfying $b\sqrt2$ being an integer, but we contradicted that by constructing a smaller one.

Answer 3

$b$ is selected to the smallest positive integer so that $b\sqrt{2}$ is an integer. One such $b$ exists by assumption, so just pick the smallest.

Next, $b^*\sqrt{2} = b(\sqrt{2}-1)\sqrt{2} = 2b-b\sqrt{2}$ is a difference of two integers, so it is an integer. Now clearly $b^* < b$ yet $b$ was supposed to the smallest $x$ with $x\sqrt{2}$ an integer, and the construction of $b^*$ contradicts this.

Answer 4

This is a bit oddly phrased. I think it would be better to write it as follows:

• Suppose $\sqrt{2}$ is rational. Let $b$ be the smallest positive integer such that $b\sqrt{2}$ is a positive integer.

  • Note the difference between this and your second [comment].

• Let $a:=b\sqrt{2}$, and let $b^*=b(\sqrt{2}-1)$. A priori this doesn't look like an integer, but it is: we have $$b^*=b\sqrt{2}-b=a-b.$$

• OK, now multiply both sides of the definition of $b^*$ by $\sqrt{2}$; this yields $$b^*\sqrt{2}=b(\sqrt{2}-1)\sqrt{2}=2b-b\sqrt{2}=2b-a.$$ Clearly $2b-a$ is an integer, so $b^*\sqrt{2}$ is an integer.

  • This addresses your fourth [comment]: the point is just that by expanding things out appropriately we can re-apply the assumption that $b\sqrt{2}$ is an integer.

• But $0<b^*<b$ (since $0<\sqrt{2}-1<1$); so $b^*$ is a positive integer smaller than $b$ which when multiplied by $\sqrt{2}$ yields an integer. This contradicts the definition of $b$.

  • ... and addresses your third and fifth [comments] - the number $b^*$ has a property it can't possibly have.

Answer 5

Usually the proof goes along this lines.

Suppose by contradiction $\sqrt{2}=\frac{a}{b}$ with $a,b\in\mathbb N$. We can assume the fraction to be reduced, in particular $a,b$ are not both even.

Then, by squaring the identity: $$2b^2=a^2$$

Thus $a^2$ is even (and so is $a$). Let $a=2c$ with $c\in\mathbb N$, and substitue in the previous. $$2b^2=4c^2$$ As before, this implies $b^2$ (and $b$) is even. Contradiction.

This seems a more straightforward proof of the irrationality of $\sqrt2$.

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As for the proof you have, you know $b^\ast$ to be a natural number since it is $a-b$. On the other hand $\sqrt{2}b^\ast=b(2-\sqrt{2})=2b-\sqrt{2}b=2b-a$.

Answer 6

Your initial premise is that $b$ is defined to be the smallest positive integer such that $b\sqrt 2$ is a positive integer. This is equivalent to reducing $\frac ab$ to its lowest terms (that is, make $a$ and $b$ coprime).

Obviously $b^\mathrm * = b\sqrt 2 - b = a-b$ will be a positive integer such that $b^*<b$.

And we now construct a new positive number $b^\mathrm {**}= b^\mathrm *\sqrt 2$. We can show this new number is an integer because $b^\mathrm*\sqrt 2 = (b\sqrt 2 - b) \sqrt 2 = 2b - b\sqrt 2 = 2b-a$.

But now we have a problem. If $b^\mathrm{**} = b^\mathrm*\sqrt 2$ is a positive integer as we've just shown, then $b$ cannot be the smallest positive integer that possesses the defining property of giving an integer when multiplied by $\sqrt 2$. We've just found an even smaller positive integer with the same property.

We have arrived at a contradiction. Therefore it is not possible to have an integer $b$ and hence $\sqrt 2$ is not rational.