# Algebra help in Hardy's 'Intro to Number Theory'

• Jul 3rd 2010, 03:43 PM
linearoperator87
Algebra help in Hardy's 'Intro to Number Theory'
Hi

I'm having trouble understanding something in Hardy and Wright's 'An Introduction to the Theory of Numbers'.

The work in question can be viewed here on Google books.

I don't understand the part beginning with 'If we divide 1 by x...'.

As best as I can figure out, we are to divide 1 by x, and then subtract 1 (that is what I think 'taking the largest possible integral quotient' means). Since x is between 1/2 and 1, 1 divided by x must be between (but not including) 1 and 2; so subtracting 1 from 1/x would leave us with just the remainder (a proper fraction) of 1/x. But 1/x - 1 = x, right? But the book says the result is 1 - x = x^2, which is not equal to x.

Any help would be great.

Thanks
• Jul 3rd 2010, 06:11 PM
chiph588@
Quote:

Originally Posted by linearoperator87
Hi

I'm having trouble understanding something in Hardy and Wright's 'An Introduction to the Theory of Numbers'.

The work in question can be viewed here on Google books.

I don't understand the part beginning with 'If we divide 1 by x...'.

As best as I can figure out, we are to divide 1 by x, and then subtract 1 (that is what I think 'taking the largest possible integral quotient' means). Since x is between 1/2 and 1, 1 divided by x must be between (but not including) 1 and 2; so subtracting 1 from 1/x would leave us with just the remainder (a proper fraction) of 1/x. But 1/x - 1 = x, right? But the book says the result is 1 - x = x^2, which is not equal to x.

Any help would be great.

Thanks

$\displaystyle \frac1x-1=x \iff x\cdot(\frac1x-1)=x\cdot x \iff 1-x=x^2$
• Jul 3rd 2010, 06:35 PM
linearoperator87
I'm okay with $\displaystyle \frac{1}{x}-1=x$ and $\displaystyle x^2=1-x$.

The problem is that the book says $\displaystyle \frac{1}{x}-1=x^2$.

Am I misinterpreting something?
• Jul 5th 2010, 05:39 PM
linearoperator87
Okay, it seems I misunderstood what the book was saying. This is what I've been told:

Given quantities $\displaystyle \alpha$ and $\displaystyle \beta$, the largest integral quotient when dividing $\displaystyle \alpha$ by $\displaystyle \beta$ is the largest $\displaystyle \gamma \in \bold{Z}$ such that $\displaystyle \beta\gamma\leq\alpha$.

So, for $\displaystyle 1$ and $\displaystyle x$ (where $\displaystyle x$ is about $\displaystyle 0.62$), the largest integral quotient is $\displaystyle 1$.

In the book, 'If we divide $\displaystyle 1$ by $\displaystyle x$, taking the largest integral quotient, viz. $\displaystyle 1$, the remainder is $\displaystyle 1 - x = x^2$,' means:

1) Find the largest integral quotient for $\displaystyle 1$ (the dividend) and $\displaystyle x$ (the divisor); the largest integral quotient is $\displaystyle 1$.

2) Multiply $\displaystyle x$ by this largest integral quotient, and subtract that from the divident (i.e. $\displaystyle 1$).

3) The remainder, then, is $\displaystyle 1 - (1\times x) = 1 - x$

Similarly for $\displaystyle x$ and $\displaystyle x^2$, $\displaystyle x^2$ and $\displaystyle x^3$, etc.