Tag Archives: proof

A minus times a minus is a plus -Are you sure you know why?

What exactly are negative numbers?
A reference , from Wikipedia:
In A.D. 1759, Francis Maseres, an English mathematician, wrote that negative numbers “darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple”.
He came to the conclusion that negative numbers were nonsensical.[25]

A minus times a minus is a plus
Two minuses make a plus
Dividing by a negative, especially a negative fraction !!!!
(10 – 2) x (7 – 3) = 10 x 7 – 2 x 7 + 10 x -3 + 2 x 3, really? How do we know?
Or we use “the area model”, or some hand waving with the number line.

It’s time for some clear thinking about this stuff.

Mathematically speaking, the only place that requires troublesome calculations with negative numbers is in algebra, either in evaluation or in rearrangement, but what about the real world ?
Where in the real world does one encounter negative x negative ?
I found two situations, in electricity and in mechanics:

1: “volts x amps = watts”, as it it popularly remembered really means “voltage drop x current flowing = power”
It is sensible to choose a measurement system (scale) for each of these so that a current flowing from a higher to a lower potential point is treated as positive, as is the voltage drop.

Part of simple circuit A———–[resistors etc in here]————–B
Choosing point A, at potential a, as the reference, and point B, at potential b, as the “other” point, then the potential drop from A to B is a – b
If b<a then a current flows from A to B, and its value is positive, just as a – b is positive
If b>a then a current flows from B to A, and its value is negative, just as a – b is negative

In each case the formula for power, voltage drop x current flowing = power, must yield an unsigned number, as negative power is a nonsense. Power is an “amount”.
So when dealing with reality minus times minus is plus (in this case nosign at all).

The mechanics example is about the formula “force times distance = work done”
You can fill in the details.

Now let’s do multiplication on the number line, or to be more precise, two number lines:
Draw two number lines, different directions, starting together at the zero. The scales do not have to be the same.
To multiply 2 by three (3 times 2):
1: Draw a line from the 1 on line A to the 2 on line B
2: Draw a line from the 3 on line A parallel to the first line.
3: It meets line B at the point 6
4: Done: 3 times 2 is 6
numberlines mult pospos
Number line A holds the multipliers, number line B holds the numbers being multiplied.

To multiply a negative number by a positive number we need a pair of signed number lines, crossing at their zero points.

So to multiply -2 by 3 (3 times -2) we do the same as above, but the number being multiplied is now -2, so 1 on line A is joined to -2 on line B

numberlines mult posneg
The diagram below is for -2 times 3. Wow, it ends in the same place.
numberlines mult posneg

Finally, and you can see where this is going, we do -2 times -3.

Join the 1 on line A to the -3 on line B, and then the parallel to this line passing through the -2 on line A:

numberlines mult negneg

and as hoped for, this line passes through the point 6 on the number line B.

Does this “prove” the general case? Only in the proverbial sense. The reason is that we do not have a proper definition of signed numbers. (There is one).

Incidentally, the numbering on the scales above is very poor. The positive numbers are NOT NOT NOT the same things as the unsigned numbers 1, 1.986, 234.5 etc

Each of them should have a + in front, but mathematicians are Lazy. More on this another day.

Problem for you: Show that (a-b)(c-d) = ac – bc – ad + bd without using anything to do with “negative numbers”

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References.
Wikipedia:
Reference direction for current
Since the current in a wire or component can flow in either direction, when a variable I is defined to represent
that current, the direction representing positive current must be specified, usually by an arrow on the circuit
schematic diagram. This is called the reference direction of current I. If the current flows in the opposite
direction, the variable I has a negative value.

Yahoo Answers: Reference direction for potential difference
Best Answer: Potential difference can be negative. It depends on which direction you measure the voltage – e.g.
which way round you connect a voltmeter. (if this is the best answer, I hate to think of what the worst answer is)
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Filed under algebra, arithmetic, definitions, education, geometrical, math, meaning, negative numbers, Number systems, operations, subtraction, teaching, Uncategorized

Euclid and vertical angles

Euclid and angle between two lines

euclid pair of lines

Euclid’s definition of angle:

From Euclid’s Elements, Book 1
Definition 8.
A plane angle is the inclination to one another of two lines in a plane which meet one another and
do not lie in a straight line.
Definition 9.
And when the lines containing the angle are straight, the angle is called rectilinear.

In the diagram we see that angle A can be taken as the inclination, but we can also see that B can be taken as the angle of inclination.

So, which is it?

If the definition is meaningful then the two angles have to be equal in size, regardless of the lack of a measurement system for angles.

My point is that the theorem about vertical angles (Euclid’s Proposition 15) is redundant, and so there is no need to prove it.

This would save students a lot of time and relieve them of the feeling that proof was pointless. This time could be better spent on proving some less obvious things.

Adding angles is a straightforward manipulative activity, but Euclid also uses subtraction of angles, which is not an obvious thing to carry out, and technically requires an additional postulate. See this:

On the formal approach to subtraction
http://aleph0.clarku.edu/~djoyce/elements/bookI/cn.html

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What is Algebra really for ?

An example tells a good tale.

Translation of a line in an x-y coordinate system:

Take a line  y = 2x -3, and translate it by 4 up and 5 to the right.

Simple approach : The point P = (2, 1) is on the line (so are some others!). Let us translate the point to get Q = (2+5, 1+4), which is Q = (7, 5), and find the line through Q parallel to the original line.  The only thing that changes is the c value, so the new equation is  y = 2x + c, and it must pass through Q.  So we require  5 = 14 + c, giving the value of c as -9.

Not much algebra there, but a horrible question remains – “What happened to all the other points on the line ?”

We try a more algebraic approach – with any old line  ax + by + c = 0, and any old translation, q up and p to the right.

First thing is to find a point on the line – “What ? We don’t know ANYTHING about the line.”

This is where algebra comes to the rescue. Let us suppose (state) that a point P = (d, e) IS on the line.

Then ad + be + c = 0

Now we can move the point P to Q = (d + p, e + q)  (as with the numbers earlier), and make the new line pass through this point:  This requires a new constant c (call it newc) and we then have  a(d + p) + b(e + q) + ‘newc’ = 0

Expand the parentheses (UK brackets, and it’s shorter) to get  ad + ap + be + bq + ‘newc’ = 0

Some inspired rearranging gives  ‘newc’ = -ap – bq – ad – be, which is equal to -(ap + bq) – (ad + be)

“Why did you do that last step ?” – “Because I looked back a few lines and figured that  (ad + be) = -c, which not only simplifies the expression, it also disposes of the unspecified point  P.

End result is:  Translated line equation is  ax + by + ‘newc’ =0,  that is,  ax + by + c – (ap + bq) = 0

and the job is done for ALL lines, even the vertical ones, and ALL translations. Also we can be sure that we know what has happened to ALL the points on the line.

I am not going to check this with the numerical example, you are !

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Congruence, proof, and rigid motions: The Common Core says WHAT, not HOW

With all the stuff in the high school geometry about proving congruence by rigid motions we get this sample geometry question from PARCC
PARCC geom test proofdef
(get the rest from numberwarrior here)

Numberwarrior’s concerns are about the language and the formal properties of congruence and I agree with him on this.
My concerns are about the stated claims of the CCSS to specify the “What do the need to know/understand/be able to do”, and the PARCC test which says “This is HOW you do a proof”.
In this particular example there are other ways of proving the assertion, not least those using the definition of congruence by rigid motions.

Let us do it this way:
1: vertical angles are equal, as there is a rotation of line AD to GC through the angle CBD, and then AD is on top of GC, so angle ABD ABF is also the angle of rotation, and is therefore congruent to angle CBD
2:There is a translation of line HE to line AD, as they are parallel. So the translation of H to H’ puts H’ on the line AD, and so angle H’BF is congruent to angle ABF.
3: But angles are preserved by rigid motions, so angle H’BF is congruent to HFG, and therefore angle ABF and HFG are congruent.

So, if I chose to teach about proof using this approach (my “HOW”) the students won’t even understand the question. Test items MUST be “Method Free”.

Also, the so called Reflexive, Symmetric and Transitive properties of congruence are no different from a=a, if a=b then b=a, and if a=b and b=c then a=c for numbers, and in both situations these are so STUNNINGLY obvious that it is cluttering up the minds of the learners to burden them with this sort of stuff. It is clear to me that this is a contribution to the CCSS from the sole pure mathematician on the committee.

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Filed under education, geometry, teaching

Geometry test. Maybe simple ! Try it.

Three maybe parallel lines, AD BE CF
Two transversals, AC DF
question how many parallel
Given that AB is equal to BC and DE is not equal to EF
(“equal” = “congruent” if you like)
show that at most two of the maybe parallel lines can be parallel.
(proof by contradiction is a last resort)

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And now Pythagoras again, with bonus

I was attempting to solve a geometrical problem the other day, a problem which, due to my complete misunderstanding, had no solution, when this popped out. It is probably bog-standard, but new to me, and this time I don’t have the heart to check if it is one of the 100 proofs of The Pythagoras theorem.

sincospythagfor blog1

Now let theta be the angle ACB. Angle ABD is then 2*theta.

Set  r = 1, then  a is  sin(2*theta)  and b is  cos(2*theta),  and so

sin(theta) = (1 – b)/a = 1/a -b/a = cosec(2*theta) – cot(2*theta)

and  cosec(theta) = (1 + b)/a = 1/a + b/a = cosec(2*theta) + cot(2*theta)

I’ve done most of the work,

so now you can show that  sin(2*theta) = 2*sin(theta)*cos(theta)

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Explain……………………..Prove

Read this, it is an excellent post on the use and meaning of the words in the hexagon (which I borrowed from Michael Pershan’s post), and other relevant words:

http://rationalexpressions.blogspot.com/2014/06/beyond-justify.html

pershan hexagon of proof

 

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-1 x -1 = 1, but some need convincing

Here is a popular argument:   -1 x -1 has to be 1 or -1

If it was equal to -1 then -1 x -1 = -1 x -1 x 1 = -1 x1 and so dividing both sides by -1 we get -1 = 1, which is not a good idea!, hence -1 x -1 = 1

This argument begs so many questions that it is difficult to know where to start.

Here is a much better one, but it does stretch the idea of area a little :

minusonetimesminusone

From the diagram  (a – 1) x (b – 1) = a x b – a – b + 1

Set a = 0 and b = 0 to get  (0 – 1) x ( 0 – 1) = 1, and since 0 – 1 is equal to -1        we get -1 x -1 = 1

This has some connection with evaluating for example  3 x ( 8 – 2)  using the distributive law.

The distributive law is a law for  a(b + c) and says nothing about  a(b – c), but never mind, we go gaily about the common task.

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Pythagoras converse, proof from scratch

There is a website with 100 proofs of the famous theorem of Pythagoras, but when I trawled the net looking for a proof of the converse, they all assume the basic theorem.

Here’s how to do it from scratch, which is considerably more satisfying, and also a simple application of similar triangles and basic algebra:

pythag converse diagram pythag converse text

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July 2, 2014 · 7:21 pm