alcazar

iTrader: (2)

What is meant by "modulus"? I can't get my head round the definition.

Noun:

• The positive square root of the sum of the squares of the real and imaginary parts of a complex number.
• A constant factor or ratio.

?????????


And how do the following work:

|-5| < 3 False.
|-15| > -12 True
|x| < 12, rewrite without the | | sig
|x-2| < or = 4 rewrite without the | | sign.

Sorry, but I've never come across "modulus" before........

Midlife......

iTrader: (2)

Modulus as in "modulus of elasticity" / Young's Modulus............the ration of stress over strain. Should be a straight line as in a constant factor or ratio which is your second definition.

Only stands for a perfect material but in real world materials have an elastic limit so is a curve.

Shaun

scubbay

iTrader: (61)

This is a different meaning.

To the OP, the modulus is essentially taking what is known as the absolute value of a number. As such it essentially ignores the sign in front of the number (- or +) and just uses the magnitude (the number itself ie 15). This is why the modulus of -15>12, because the modulus of -15 is 15 and as such is greater than 12.

Hope this helps and if not let me know.

scubbay

iTrader: (61)

Therefore in response to your problems above:

l-5l = 5 which is greater than 3. Therefore that statement is false. Apply the same logic to the similar questions.

For the others:

lxl < 12. Therefore -12< x <12. This means that the modulus of any number in that range will be less that 12. The lower limit has to be added as if x=-13 for example its modulus would be 13 and as such not less than 12 so the above defines the finite range x can lie in.

lx-2l <=4 can be rewritten as x-2<=4. This means x<=6. Now x could also be equal to a number equal to or larger than -2 as this would also satisfy this equation. Now its been a long time since I have done this so dont know how I would present this solution properly. I guess you state that x-2<=-4 as this would also satisfy the modulus criteria.

Therefore the answer is -2< x < 6.

speedking

-2<=x<=6

MOD(X) = ABS(X)

scubbay

iTrader: (61)

You are right, I missed the = part in my answer.

alcazar

iTrader: (2)

Now that makes more sense, thanks Scubbay, but the question arises, what is it FOR?

scubbay

iTrader: (61)

Well I can only speak from an engineering sense where the absolute value or modulus of a number is used in order to get things like the magnitude of stresses regardless of there state ie +ve or -ve. From a Maths perspective it is a way of getting around taking the square root of a -ve number, you say the square root of modulus of x is as that means you can define it as its real part only (complex number is the real root of a negative number and will have a real and imaginary part).

alcazar

iTrader: (2)

Thankyou especially scubbay.

And the engineering bit makes sense, he's on an engineering course.

CREWJ

We use it quite a lot in work when comparing stresses or load cases when we want to automate calculations related to the magnitude rather than the vector.

Leslie

Physics and Chemistry were really interesting, but I found the Maths was very boring. I think the teacher who was brought in out of retirement and was good at maths but incompetent when it came to teaching it was the source of the difficulties.

Les

alphaj12

Pure Maths is just unreal

Anything past o level/gcse does not make any sense

alcazar

iTrader: (2)

I have to agree: I used to ask my maths teacher, forty years ago, "But miss, what's it FOR???????"

CREWJ

I was presenting at an event on Friday trying to entice kids to do engineering. We just presented some of the "pointless" maths and showed where it would be applicable.

scud8

iTrader: (1)

We probably wouldn't have Scoobynet if the use of abstract Hilbert Spaces to calculate quantum mechanical effects hadn't led to the current understanding of semiconductors.

alcazar

iTrader: (2)

Oh aye.....but FORTY YEARS AGO?

Saint AAI

|x| < 12, rewrite without the | | sig

sqrt(x^2) < 12 as when you square a number then square root it, you get rid of the negative value, if it has one. Except if you ever see i^2 or j^2, then it equals -1 as i or j is the imaginary part. I use j as I'm an electrical engineer, but I think mathematicians use i.

Using the absolute value of something is also used in statistics to calculate average deviation and probably also standard deviation.

alcazar

iTrader: (2)

Ah..........standard deviation, apparently that's coming up soon........

CREWJ

That's easy. It's just how much something varies. It's used in Six Sigma in the real world.

Think of it this way, if a part is made to a specific size then it will vary from part to part. Standard deviation is used to calculate how much that size will vary.

finalzero

Strange as it sounds, I am a Software Engineer but I am not good at Maths..."Calculator Kid of the 21st Century".

I am due to get through and learn it properly now, pure maths is amazing, in fact the whole range of Mathematics is programming at it's purest form.