Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one quick challenge before we start. I will read one statement, and you tell me: true or false? If z squared is real, then z is real. Think for a moment. True or false? Most of you said true, right? It sounds so obvious. Hold on to your answer. Look, JEE has asked one line statements like this: they look obvious, but one small word can hide a trap. Today we learn to settle them in one line. So, let's begin.

0:34Part C: true or false, in one line

So, what is Part C? Each item is one statement, and you decide: true or false. But a guess is not enough, okay? You must give a one line reason, or a counterexample. And what exactly is a counterexample? One single case where the statement fails. Now think. Suppose a statement claims that something always happens. To prove it false, how many failing cases do you need? Just one! Yes, exactly. One counterexample is enough. If it fails even once, the statement is false, that's it. But true is harder. Checking two or three nice values proves nothing, okay? For true, you need a one line reason that works in every case. Clear?

1:17Practice: the cube roots of unity

Alright, let's practise. Look at the left of the figure: the cube roots of unity, one, omega and omega squared. Omega is this Greek letter, okay? Statement: the cube roots of unity form an equilateral triangle. The picture already says triangle, right? But a picture is not a reason. So, true or false, and what is your one line reason? True! And the reason in one line: all three lie on the unit circle, one hundred and twenty degrees apart. The small circle after the number means degrees. Equally spaced points on a circle always make a regular figure, so the reason covers every case. On the right, the same idea for the n-th roots, with n equal to eight. They are labelled alpha, alpha squared, alpha cubed, and each one is two pi by n further round the circle, so they make a regular eight sided figure. Got it? Now the next one. Omega to the power n, plus omega to the power two n, equals minus one, for every integer n. Put n equal to one: omega plus omega squared, and you know that is minus one. Looks true, right? But for every integer? No, no, careful! It is false. Omega cubed is one. So take n a multiple of three; this vertical line means three divides n. Then omega to the n is one, and omega to the two n is also one. The sum is one plus one, two, not minus one! One counterexample, and the statement is gone. See, the trap was hiding in the word every.

2:51Example 1: true or false on z^2, z bar and arguments

Right, on to Example one, and our puzzle is the very first statement. One: if z squared is real, then z is real. Two: if mod z equals one, then z bar equals one by z. These two vertical bars mean mod, the modulus, and z bar, the z with a bar on top, is the conjugate, okay? Three: arg of z one z two equals arg z one plus arg z two, for principal arguments. The small one and two below are just labels. Four: mod z is less than or equal to mod of R e z, plus mod of I m z. Here R e z means the real part of z, and I m z the imaginary part. True or false, each with a reason. Pause the video, and try. Done? Okay, let's check. Statement one, our puzzle. Take z equal to iota, the imaginary unit. Iota squared is minus one, which is real. But iota itself is not real! One counterexample, so it is false. Said true at the start? No problem, most students do. Look, that was the trap. Statement two. Here mod z is one. Now z times z bar is mod z squared, which is one. The double arrow means implies. Divide by z, and z bar equals one by z. True! With me? Statement three looks like a rule you know, right? Arg of a product is the sum of the args. But read the last words: for principal arguments. And the principal argument lies between minus pi and pi. The round bracket means minus pi is left out, and the square bracket means pi is included. So, still true? False! Take z one and z two both equal to minus one. The principal argument of minus one is pi, so the right side is pi plus pi, two pi. But z one z two is plus one, and arg of one is zero. Zero is not two pi; two pi is outside the principal range. Got it? Last one, statement four, and this one is true. Let's see why. Write z as x plus iota y. Then mod z is root of x squared plus y squared, the real part is x, and the imaginary part is y. So the claim is: root of x squared plus y squared is at most mod x plus mod y. Square both sides. Left: x squared plus y squared. Right: x squared plus y squared, plus two mod x mod y. That extra term is never negative, so the inequality always holds, for every z. True! So, two true and two false. All clear?

5:32Example 2: true or false on quadratics and inequalities

Now Example two, a little harder: four statements on quadratics. One: x squared plus x plus one is greater than zero, for all real x. Two: if a is positive and D, the discriminant, is negative, then a x squared plus b x plus c is negative for some x. Three: the minimum of x squared minus four x on the interval three to five is minus four. The square brackets mean both ends are included. Four: x minus one, whole squared, times x minus two, is greater than or equal to zero, if and only if x is at least two. The double headed arrow means if and only if. Pause the video and try. Careful with every word! Done? Okay, let's check. Statement one: a is one, positive, and D is one minus four, minus three, negative. So the parabola opens upwards and never meets the x axis. It stays above it for every real x. True! Statement two is the same situation, in general. When the coefficient a is positive and D is negative, the expression is always positive, so it is never negative. False! In fact, statement one was just a special case of this. Now statement three. Call the expression f of x. Most students do this: the vertex is at x equals two, and f of two is four minus eight, minus four. So the minimum is minus four. True, right? No, no, careful! Read the interval again. The vertex, x equals two, is not inside three to five. This symbol, the curly e with a slash through it, means not in. So the value minus four is never reached here. False! Then where is the minimum? The parabola opens upwards, and the whole interval lies to the right of the vertex, so the graph keeps rising there. The lowest point is at the left end: f of three, nine minus twelve, minus three. Really simple, isn't it? Last one, statement four. Hmm, x minus one, whole squared, is never negative. So we only need x minus two at least zero: x at least two. True, right? Wait, what about x equals one? Then the square is zero, and zero times minus one is zero. Zero is greater than or equal to zero, so x equals one also works! But one is not at least two, so the if and only if breaks. False! Whenever a factor is squared, check the point where it becomes zero. Okay?

8:10Quick revision: counterexample or a one-line reason

So, quick revision. False needs just one counterexample. True needs a one line reason that works in every case. And read every word: every, principal, the interval, if and only if. That is where the traps hide. Now try the other concept checks on this page, and say your answer and your reason before you open the solution. All the best!