Via @DrTrapezio, an interesting question: Year 13 curve sketching challenge: y = | | | |x| - 1| - 1| - 1| — Luciano Rila (@DrTrapezio) July 14, 2016 Where do you even start with that mess? The answer to that, my friend, is you start in the middle and

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Posted in ask uncle colin, modulus.

Dear Uncle Colin, I have to solve the inequality $x^2 - \left|5x-3\right| \lt 2+x$. I rearranged to make it $x^2 - x - 2 \lt \left|5x-3\right|$ , but the final answer is eluding me. -- Put Right Inequality Muddle Hello, PRIM! You're off to a good start; the next thing

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Posted in ask uncle colin, modulus.

Dear Uncle Colin, I have an inequality that involves two pairs of modulus signs, and I can't make head nor tail of it. It's $\left| \left| 3 - x \right| - 2x\right| \le 9$. Please help! -- Absolutely Bewildered, Driven Absolutely Barmy Well, ABDAB, I'd normally recommend sketching something like

Read More →You know the ridiculous kind of pseudo-context question that makes you go 'Why doesn't Lisa get a proper hobby rather than timing her friends doing jigsaw puzzles?'? You could replace pretty much all of them with "A mathematician is trying to be clever by...". In this particular case, a mathematician

Read More →It’s usually quite simple to spot the error in ‘proofs’ that $1=2$: either someone’s divided by 0 or glossed over inverting a multi-valued function (conveniently forgetting the second square root, for example). You sometimes (as with the sum of natural numbers being $-\frac{1}{12}$, if you throw out all good sense)

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Posted in core 3, inequalities, modulus.

Solve $\sqrt{|x|-3} > x-4$ Difficult, as the man said. Difficult, difficult, lemon difficult. It’s not that it’s tricky to solve it - it’s just… fiddly. Let’s start by drawing the graphs. The right-hand side is easy enough: it’s a straight line with gradient 1, through the point $(0,-4)$. The left-hand

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