How to calculate the square or cubic root of a number on a simple calculator?

[JeffM]

I meant circular in the English sense of the word. Trying to define something, and then using the same something in the definition, therefore not really explaining what it is or how it truly works.

But that is not how the definition works. [imath]a^0[/imath] is not defined with respect to any power. The integer powers greater than 0 are explained in terms of lower powers, which eventually get you back to that trivially but clearly defined [imath]a^0 = 1.[/imath]. The English phrase "circular definition" means explaining a thing in terms of itself, which is never done in a recursive definition. It is, if you like, a procedural definition, but it does not have the logical flaw of implied by circularity because it terminates.

Logically, it is exactly like a function call that calls itself only a finite number of times. There is of course a flaw if the process may not terminate. There is no flaw if the process must terminate after a finite number of iterations.

 

[Deleted member 4993]

Logically, it is exactly like a function call that calls itself only a finite number of times.

Like Factorial

 

[JeffM]

Like Factorial

Indeed.

 

[Steven G]

I disagree. Multiplication is repeated addition (fractions are commonsense):
3*2 = 3+3 = Two 3s
3*1 = 3 = One 3s
3*0.5 = 1.5 = Half 3s

Looking at multiplication as an entirely unique concept that has nothing to do with addition is way more confusing, and definitely wrong.

Yes, in your examples multiplication is repeated addition. However, as Dr Khan is trying to point out, is 4.2*7.8 a repeated addition problem?
I am in favor of saying that multiplication is repeated addition, but only initially. A perfect example is learning the times tables. If you know that 7*6 =42, then 7*7 = 7*6 + 7. Why? 7*7 = 7+7+7+7+7+7+7 = (7+7+7+7+7+7)+7 = (7*6) + 7 = 42 +7 = (after counting on your fingers) 49.

 

[JeffM]

As always, the real difficulty begins with the real numbers (word play intentional).

Obviously, we can start by defining multiplication as repeated addition on the non-negative integers.

[math]\text {Given } a \text { whole number} \ge 0 \text { and } b \text { a whole number} \ge 0, \text { then}\\ a \times b = 0 \text { if } b = 0, \text { and }\\ a \times b = a + a \times (b - 1) \text { if } b > 0.[/math]
And, without the notation, it is probably the easiest way to introduce multiplication to kids under 8. Moreover, you can extend the definition to multiplication of negative integers and rationals generally. Whether extending that definition is is the best method pedagogically, I have no clue. What is interesting to me is that you can take the method backwards to define addition recursively in terms of the successor function, which actually connects grade school arithmetic to the foundations of mathematics.

 

[lookagain]

I disagree. Multiplication is repeated addition (fractions are commonsense):
3*2 = 3+3 = Two 3s
3*1 = 3 = One 3s
3*0.5 = 1.5 = Half 3s

Actually, in a limited sense (I'm thinking arithmetic in elementary school for an example), if the first number is a positive whole number, and the second number
is a non-negative whole number, fraction, or decimal number, then your examples
would be interpreted as:

3*2 is three twos, which equals 2 + 2 + 2
3*1 is three ones, which equals 1 + 1 + 1
3*0.5 is three "times" 0.5 (or three copies of 0.5), which equals 0.5 + 0.5 + 0.5

 

[Apple30]

I'm even more confused now than I was before. What happens if I go to an exam where calculators are not allowed, and they tell me to approximate on paper 2.1^0.4? How would I do something like that?

 

[JeffM]

I'm even more confused now than I was before. What happens if I go to an exam where calculators are not allowed, and they tell me to solve on paper 2.1^0.4?

Would this be impossible to solve on paper? Surely it can't be...

I’d probably go for Newton’s method and an initial guess of 1.

 

[Apple30]

I’d probably go for Newton’s method and an initial guess of 1.

Could you please direct me to instructions or video of this?

 

[Dr.Peterson]

I'm even more confused now than I was before. What happens if I go to an exam where calculators are not allowed, and they tell me to approximate on paper 2.1^0.4? How would I do something like that?

Do you have reason to think they would actually do that to you? Have you ever been given such an assignment, or are you just imagining the possibility?

Generally, on no-calculator exams, we give problems that make sense to be solved without calculators. In the days before calculators, if there was a need to do such a calculation, you would be given logarithm tables (which were the way to do these things before calculators). The only reason I can imagine for asking for 2.1^0.4 on an exam would be if they had taught you a way to do it, and want to see if you learned. In real life, you would never do this by hand.

 

[Apple30]

Do you have reason to think they would actually do that to you? Have you ever been given such an assignment, or are you just imagining the possibility?

Generally, on no-calculator exams, we give problems that make sense to be solved without calculators. In the days before calculators, if there was a need to do such a calculation, you would be given logarithm tables (which were the way to do these things before calculators). The only reason I can imagine for asking for 2.1^0.4 on an exam would be if they had taught you a way to do it, and want to see if you learned. In real life, you would never do this by hand.

The main reason why I ask is because, I feel like my brain cannot learn something unless I can see how it works. With programming, functions are made of simpler functions, which are made of operations, which are made of assembly, which are made of logic gates. So everything can be broken down into simple and fundamental building blocks.

But with mathematics, I just feel like everything is a black box, where I just have to trust the "rules" and trust the "calculator". I can certainly get the answer, but I still don't "understand" what any of this means. If I don't understand how powers work, how can I possibly understand more advanced math concepts that depend on powers?

I think the education system is really broken in a way. In the old days people had no choice but to do everything on paper, and there were also illustrations and context. Nowadays, it's just rules and calculators, without understanding.

Also, I should point out, that I am still in the process of learning arithmetic, as I am horribly slow at it. I don't know anything about "logarithms", and maybe that could be a reason why the concept of powering decimals don't make sense? What should I learn after arithmetic? I really want to learn maths in the correct order so that I can actually understand this stuff.

 

[Steven G]

You are correct that you should not accept any rules. You need to understand the rules. I have always told my students not believe anything that I tell them. I suggested that they go home and convince themselves that I was correct.

 

[Apple30]

Incorrect

π * √2 ~ 4.442883

That is what I get from Excel. Whom should I believe - you or Bill Gates?

Yes, sorry you're right Sabhotosh, my apologies. It's hard to break away from the many years of this religious brain washing "multiplication is series of addition".

I suppose maybe I should look at it like: some operators work one way for whole numbers, and another way for decimal numbers. Other operators may work differently for negatives vs positives. Addition becomes like-subtraction when the numbers are negative. Multiplication works like-division when decimals are used? Kind of makes me wonder why this is? Why does an operator "change" depending on the number?

 

[Dr.Peterson]

The main reason why I ask is because, I feel like my brain cannot learn something unless I can see how it works. With programming, functions are made of simpler functions, which are made of operations, which are made of assembly, which are made of logic gates. So everything can be broken down into simple and fundamental building blocks.

But with mathematics, I just feel like everything is a black box, where I just have to trust the "rules" and trust the "calculator". I can certainly get the answer, but I still don't "understand" what any of this means. If I don't understand how powers work, how can I possibly understand more advanced math concepts that depend on powers?

This is a very different question than wanting to be able to solve problems on a test without a calculator!

I think the education system is really broken in a way. In the old days people had no choice but to do everything on paper, and there were also illustrations and context. Nowadays, it's just rules and calculators, without understanding.

Also, I should point out, that I am still in the process of learning arithmetic, as I am horribly slow at it. I don't know anything about "logarithms", and maybe that could be a reason why the concept of powering decimals don't make sense? What should I learn after arithmetic? I really want to learn maths in the correct order so that I can actually understand this stuff.

It's good to want to fully understand what you are doing; but in fact, doing everything on paper doesn't accomplish that.

"In the old days" they didn't understand better; most students were just taught mere rules, things to do on paper without knowing why! It is today's education (at least in America, and at least in theory) that is trying to avoid mere rules, and teach for understanding. (I don't think they do well at that, in part because most elementary teachers themselves lack deep understanding. But also, parents resist change, because they don't understand!)

Logarithms are a tool (invented in the 1600s) to make calculations on paper easier. A few people spent years doing paper calculations to create tables, which you would then use to convert numbers to their logarithms, which you could add, in order to multiply the original numbers, or multiply by n to raise the original number to the nth power, and so on. When I was in school, we would use those tables (and also tables of trigonometric functions, etc.), or a slide rule, which uses the same concepts in a physical way to do the calculations.

Understanding doesn't really help with calculation. (Just a little.) What it helps with is choosing the right calculations to do, and knowing how to check your answers. For example, I could use the various methods that have been suggested (without needing to think about why the methods worked!) to calculate 2.1^0.4 from scratch, if I were willing to put in a lot of work, and get a few digits of accuracy. But understanding would come into play when I decided to check my work by raising my answer to the 5th power, and comparing that to the square of 2.1 (and knowing how much accuracy to expect).

 

[JeffM]

You asked about Newton’sMethod

https://www.bing.com/videos/search?q=video+newton’s+method&&view=detail&mid=D383339C061C347D336FD383339C061C347D336F&rvsmid=B6AA90336651FD95C012B6AA90336651FD95C012&FORM=VDQVAP

You asked about the results of mathematical reasoning.

The rules of arithmetic were treated for many millennia as what today would be called scientific laws: facts that are derived from experiment or observation. (A philosopher of science such as Popper would say my formulation scientific law is too crude, but the “fact” that 2+4=6 was deduced from experience long before any philosopher of science existed.) Scientific laws will be rejected if confronted with contrary evidence. (Another over-simplification.)

The Greeks were the first culture known to have insisted on explicit reasoning to reach general conclusions about mathematics. However, they recognized that you cannot make something from nothing. They appear to have invented the axiomatic method of constructing by logical reasoning a large body of mathematical results from a small number of assumptions (axioms). I suspect that they viewed their axioms as scientific laws in the crude sense mentioned above.

The inability to put calculus on such a footing and the discovery that Euclidian geometry did not rest on axioms that are necessarily true of the physical world changed the way mathematicians think. Most mathematicians now dismiss the notion that mathematics is a science based on observation or experiment. The results of correct reasoning from axioms are immutable, but the axioms are purely arbitrary provided they do not lead to contradiction. That viewpoint led to much of the flavor of modern mathematics such as the way mathematicians define things.

To develop arithmetic and algebra, you need a fairly extensive set of axioms, the axioms of an ordered field. Those axioms can in fact be derived from a simpler set of axioms, but, and this is a personal view, you can also justify the field axioms as scientific laws in the modern sense.

 

[Cubist]

The result of calculating a root is often an irrational number (it can only be approximated as a decimal number). Therefore, particularly in maths, such numbers are usually left written as roots. Therefore [imath]\sqrt{2}[/imath] will just be left as [imath]\sqrt{2}[/imath]. (Engineers would be more likely to "dive in" and replace it with an approximation like 1.414)

Don't memorise this. It's more advanced than your current level. If you're interested you could write a program to do this. It might be similar to the method your computer/ calculator uses to calculate powers. With that caveat, here's how to find 2.1^0.4 using Newton's method...

Let \(\displaystyle x = N^{0.4} \implies x = N^{2/5} \implies x^5=N^2\)

Use this function that gives 0 when x is correct...
\(\displaystyle f(x) = x^5 - N^2\)
Differentiate...
\(\displaystyle f'(x) = 5x^4\)

Newton's method...
\(\displaystyle x_{n+1} = x_n - \frac{f( x_n )}{f'( x_n )}\)
\(\displaystyle x_{n+1} = \frac{5x_n^4 \times x_n}{5x_n^4} - \frac{x_n^5 - N^2}{5x_n^4}\)
\(\displaystyle x_{n+1} = \frac{4x_n^5 + N^2}{5x_n^4}\)
Let N=2.1...
\(\displaystyle x_{n+1} = \frac{4x_n^5 + 4.41}{5x_n^4}\)

Start with a guess of 1.5, and the following red digits differ from the actual answer...
[imath]x_1 = 1.5\\ x_2 = 1.3\red{7422222222222222}\\ x_3 = 1.34\red{668723061614619}\\ x_4 = 1.34551\red{448938036008}\\ x_5 = 1.34551244150\red{792604}[/imath]

 

[Deleted member 4993]

The result of calculating a root is often an irrational number (it can only be approximated as a decimal number). Therefore, particularly in maths, such numbers are usually left written as roots. Therefore [imath]\sqrt{2}[/imath] will just be left as [imath]\sqrt{2}[/imath]. (Engineers would be more likely to "dive in" and replace it with an approximation like 1.414)

Don't memorise this. It's more advanced than your current level. If you're interested you could write a program to do this. It might be similar to the method your computer/ calculator uses to calculate powers. With that caveat, here's how to find 2.1^0.4 using Newton's method...

Let \(\displaystyle x = N^{0.4} \implies x = N^{2/5} \implies x^5=N^2\)

Use this function that gives 0 when x is correct...
\(\displaystyle f(x) = x^5 - N^2\)
Differentiate...
\(\displaystyle f'(x) = 5x^4\)

Newton's method...
\(\displaystyle x_{n+1} = x_n - \frac{f( x_n )}{f'( x_n )}\)
\(\displaystyle x_{n+1} = \frac{5x_n^4 \times x_n}{5x_n^4} - \frac{x_n^5 - N^2}{5x_n^4}\)
\(\displaystyle x_{n+1} = \frac{4x_n^5 + N^2}{5x_n^4}\)
Let N=2.1...
\(\displaystyle x_{n+1} = \frac{4x_n^5 + 4.41}{5x_n^4}\)

Start with a guess of 1.5, and the following red digits differ from the actual answer...
[imath]x_1 = 1.5\\ x_2 = 1.3\red{7422222222222222}\\ x_3 = 1.34\red{668723061614619}\\ x_4 = 1.34551\red{448938036008}\\ x_5 = 1.34551244150\red{792604}[/imath]

I don't know anything about "logarithms",

@Apple30 had indicated that s/he does not know manipulation of logarithm. I assume calculus not in student's repertoire. Thus employing differentiation would be very difficult if not impossible.

 

[JeffM]

Yes, sorry you're right Sabhotosh, my apologies. It's hard to break away from the many years of this religious brain washing "multiplication is series of addition".

I suppose maybe I should look at it like: some operators work one way for whole numbers, and another way for decimal numbers. Other operators may work differently for negatives vs positives. Addition becomes like-subtraction when the numbers are negative. Multiplication works like-division when decimals are used? Kind of makes me wonder why this is? Why does an operator "change" depending on the number?

A few points.

Mathematicians work with many different kinds of numbers. It is easiest to start with natural numbers defined as 0, 1, 2, 3 etc. And mathematicians work with many kinds of operations. The ones familiar from grade school are addition, subtraction, multiplication, and division.

We notice that addition has the opposite effect as subtraction. We call these operations inverse operations meaning that they cancel each other. 6 - 4 + 4 = 6. And we notice that multiplication has the opposite effect as division. Multiplication and division cancel each other. 6 [imath]\div[/imath] 3 [imath]\times[/imath] 3 = 6.

A property that is useful is called closure. If I add ANY two natural numbers together, I get a natural number. If I multiply ANY two natural numbers together, I get a natural number. Subtraction and division, however, do not have this nice property of closure in the natural numbers.

So we expand our number system to include negative numbers and call the expanded system the integer numbers. When we do that, we get closure for subtraction. Nice. But we want to keep as much as possible that is true of the natural numbers true for the integers and also want to make the integer numbers useful. This may call for adjusting the rules on how addition and subtraction are computed, but the adjusted rules must give the same answers when dealing only with natural numbers.

And we can expand the integer system to a new system called rational numbers where division of any rational number by any rational number except zero is a rational number. Now we have closure for division (except division by zero). Again, we have to adjust the rules for how addition, subtraction, multiplication, and division are computed, but the adjusted rules must give the same answers when dealing with only with integers or natural numbers.

Something interesting happens when we get to the rational numbers; it turns out that, in principle at least, we do not ever need subtraction and division. We can do all our conceptual work with just addition and multiplication.

Now the mathematicians have kept on doing this and come up with the real numbers, the complex numbers (both of which have no easy explanation but are handy for calculus), and even weirder things like quaternions (which are handy for computer graphics).

The principle to remember is that though expanded systems have adjusted rules for computing answers, those rules give the same result as the rules you learned in grade school when applied to natural numbers.

 

[Cubist]

@Apple30 had indicated that s/he does not know manipulation of logarithm. I assume calculus not in student's repertoire. Thus employing differentiation would be very difficult if not impossible.

Yes, I knew that, that's why I wrote...

It's more advanced than your current level...

But sometimes there's no harm having a small glimpse into "the future"

We could also get the exact same result without calculus by...

Let x = u + v
(u + v)^5 = N^2

Expand the bracket on the LHS (binomial expansion)...
u^5 + 5*u^4*v + 10*u^3*v^2 + 10*u^2*v^3 + 5*u*v^4 + v^5 = N^2
u^5 + 5*u^4*v ≈ N^2 for small v because v^2, v^3 and v^5 all become MUCH smaller and can be approximated as zero
u^4 * (u + 5*v) ≈ N^2
u + 5*v ≈ N^2/u^4
v ≈ (N^2/u^4 - u)/5

Let [imath]u = x_n[/imath] be our guess. A better guess is obtained by calculating v (above) and the next guess becomes u+v

[imath]x_{n+1} = x_n + (N^2/x_n^4 - x_n)/5[/imath]

This can be rearranged as...

\(\displaystyle x_{n+1} = \frac{4x_n^5 + N^2}{5x_n^4}\)

which bypasses the calculus step

 

[Apple30]

Thanks everyone for your kind support, and yes, it is definitely way beyond my capability. I am still learning basic arithmetic
I guess the reason why I originally asked this was because I assumed that after I finish learning basic arithmetic I would be ready to move on to powers, which I assumed would be a simple concept, but I guess not...

I'm probably going to need to buy a large set of math "curriculum" books, that go all the way from beginner to advance. Since I am an adult, I really cannot go back to primary school.

I would appreciate any advice on what books I should buy, so long as it comes as a set for consistency reasons, and is easy to understand. I am aware of online places like Khan and Brilliant, but I would also like something physical that I can flick through and then do worksheets.

"In the old days" they didn't understand better; most students were just taught mere rules, things to do on paper without knowing why!

Well, I assumed things were different because of this video.