# A problem from Hardy's_Pure Mathematics on Real variables about Linear equations in 3 variables

It may be my stupidity, yet I am unable to understand what the following problem asks
(Q.1 Miscellaneous Examples, Chapter-1: Real Variables).

It says:

What are the conditions that $$ax+by+cz=0$$,

• (1) for all values of $$x,y,z$$ ;
• (2) for all values of $$x,y,z$$ subject to $$\alpha x+\beta y+\gamma z=0$$;
• (3) for all values of $$x,y,z$$ subject to both $$\alpha x+\beta y+\gamma z=0$$ and $$Ax+By+Cz=0$$?"

Initially I thought it was about linear diophantine equations in three variables, but then it is already given that the constant term is $$0$$ which means that it is a solvable diophantine equation (for all the coefficients divide $$0$$). So the problem is surely not about the solvability of the equation. Then, what it is about? What does the author mean by 'conditions'? Any suggestions are welcome.

• What are $a,b,c,x,y,z$??? – amsmath Mar 23 at 5:20
• But Diophantine equations only involve integers... – amsmath Mar 23 at 6:08
• Very poorly written question, so not your fault. – user21820 Mar 23 at 11:12
• @user21820 It was perfectly clear and well written following the conventions used when Hardy wrote it. It's only "poorly written" in the sense that "Shakespeare is poorly written, because he couldn't even spell many of the words correctly according to a 21st-century dictionary". – alephzero Mar 23 at 14:17
• @alephzero: Your claims are unjustified. – user21820 Mar 23 at 14:57

I think Hardy is asking for conditions on $$(a,b,c)$$, to be given in terms of $$(A,B,C)$$ and $$(\alpha,\beta,\gamma)$$. The $$(x,y,z)$$ are dummy variables, note the "for all".

For (1), which $$(a,b,c)$$ are orthogonal to all vectors? Answer, clearly only the zero vector, so $$(a,b,c)=(0,0,0)$$.

For (2), which $$(a,b,c)$$ are orthogonal to all vectors which are orthogonal to $$(\alpha,\beta,\gamma)$$? Answer, clearly those which are parallel to $$(\alpha,\beta,\gamma)$$. So the condition is that for some scalar $$\lambda$$, we must have $$(a,b,c)=\lambda (\alpha, \beta, \gamma)$$.

Case (3) is left to the reader; the condition is clearly that for some scalars $$\lambda, \mu$$ we must have that $$(a,b,c)=\lambda (\alpha, \beta, \gamma)+\mu (A,B,C)$$.

• Geometrically, the solution to (1) is the origin. The solution to (2) is the line through the origin and $(\alpha, \beta, \gamma)$. And the solution to (3) is the plane containing the origin, $(\alpha, \beta, \gamma)$ and $(A,B,C)$. – Paul Sinclair Mar 23 at 13:57

Since I don't have the book available, nor do I recall ever reading it (at least not any recent version of it), I can't be sure what it is getting at as I don't have the context of knowing what was written in Chapter $$1$$. It should likely help give you an idea.

However, here is what it seems to me. As it's a book about real variables, I doubt the equations are supposed to be diophantine ones, i.e., where the variables can only be integers. Instead, it's to do with the degrees of freedom vs. how many linearly independent constraints you have. With just $$1$$ equation, you can generally pick any $$2$$ values and then determine the third, e.g., choose $$y$$ and $$z$$ to then get $$x$$ from

$$x = \frac{-by-cz}{a} \tag{1}\label{eq1}$$

This is assuming that $$a \neq 0$$.

With $$2$$ constraint equations, assuming they're linearly independent, you can solve each for $$1$$ variable to eliminate it, resulting in an equation with $$2$$ variables. You then have $$1$$ degree of freedom to choose $$1$$ of those remaining variables, with the other $$2$$ then being determined. For example, using your provided $$2$$ equations:

$$ax + by + cz = 0 \tag{2}\label{eq2}$$ $$\alpha x + \beta y + \gamma z = 0 \tag{3}\label{eq3}$$

Multiply \eqref{eq2} by $$\alpha$$, multiply \eqref{eq3} by $$a$$, and subtract the $$2$$ resulting equations to get

$$\left(\alpha b - a\beta\right)y + \left(\alpha c - a\gamma\right)z = 0 \; \Rightarrow \; y = \left(\frac{a\gamma - \alpha c}{\alpha b - a\beta}\right)z \tag{4}\label{eq4}$$

This is assuming that $$\alpha b - a\beta \neq 0$$. Thus, for each $$z$$, you get a $$y$$ from \eqref{eq4} and then an $$x$$ from \eqref{eq1}.

For $$3$$ linear constraint equations, assuming they're all linearly independent, you get just one set of values.

If any of these equations are linearly dependent, this results in an extra degree of freedom for each equation that is dependent.

Based on your understanding of chapter 1, please indicate if this doesn't match what you might expect.

Update: Based on the comment by ancientmathematician about a geometric argument, note that one equation gives a plane, having $$2$$ independent equations gives the intersection of $$2$$ planes, i.e., a line, and $$3$$ independent equations gives the intersection of $$3$$ planes, i.e., a point. Perhaps these are the "conditions" being referred to. However, based on how the question is specifically worded, I think a better fit would be what is described in ancientmathematician's answer.

• the 1st chapter is about rational and irrational numbers, irrationality, surds, linear arithmetic continuum, Dedekind theorem, Weierstrass theorem and limit points. – Awe Kumar Jha Mar 23 at 6:31
• @AweKumarJha Thanks for the info. Offhand, I don't see any direct connection between the problem and any of those topics, but linear arithmetic continuum seems like the closest one. Do you think what I wrote fits to at least some extent with what the problem is asking for? – John Omielan Mar 23 at 6:35
• in fact what you wrote is the only thing that presently fits to a remarkable extent to the context, for I don't see other possibilities. Anyway, thanks for your support. – Awe Kumar Jha Mar 23 at 6:41
• In the 1921 3rd edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 31, and in the 1952 10th edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 32. Incidentally, I see nothing in Chapter I to suggest that this has anything to do with Diophantine equations. Indeed, about the only thing in this chapter that I can find even remotely involving Diophantine equations are some irrationality proofs by contradiction. – Dave L. Renfro Mar 23 at 7:16
• I rather suspect that Hardy (and those for whom he wrote the book) would have given a geometric and not algebraic argument. I don't know why the problem is set here in the book, but of course it's the sort of thing needed when one comes to Lagrange multipliers. – ancientmathematician Mar 23 at 7:57