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

[Apple30]

Is it possible? For example, I know we can square (n*n), or cube (n*n*n) using basic arithmetic multiplication. With an advanced calculator we could find the root using the power symbol like n^(1/2) for square roots, and n^(1/3) for cubic roots, but if a calculator doesn't have a square or power button, is it still possible?

I'm having a hard time understanding in layman terms, how to make sense of "Multiplying a number by itself decimal-number of times". What does that even mean?

Can someone break this down for me, in layman terms and basic arithmetic?

Thanks in advance.

 

[Deleted member 4993]

Is it possible? For example, I know we can square (n*n), or cube (n*n*n) using basic arithmetic multiplication. With an advanced calculator we could find the root using the power symbol like n^(1/2) for square roots, and n^(1/3) for cubic roots, but if a calculator doesn't have a square or power button, is it still possible?

I'm having a hard time understanding in layman terms, how to make sense of "Multiplying a number by itself decimal-number of times". What does that even mean?

Can someone break this down for me, in layman terms and basic arithmetic?

Thanks in advance.

Read:

The Root of the Matter: Approximating Roots with the Greeks -

The Root of the Matter: Approximating Roots with the Greeks

​

 

[Deleted member 4993]

I'm having a hard time understanding in layman terms, how to make sense of "Multiplying a number by itself decimal-number of times". What does that even mean?

If we have ........................ A * B/C ................................ (assume B and C are integers)

and if A * B = D (another integer), then,

A * B/C = (A * B)/C = D/C

So if you have the number say (2/3), which you want to multiply by itself, then

(2/3) * (2/3) = (2*2)/(3*3) = 4/9

 

[Deleted member 4993]

but if a calculator doesn't have a square or power button, is it still possible?

If your calculator has a "log" and/or "ln" button, you can convert the "root" problem to a "multiplication/division" problem and continue. But that is NOT basic arithmetic.

Easiest thing to do would be to buy a new calculator with a [x^y] button.

 

[Apple30]

If we have ........................ A * B/C ................................ (assume B and C are integers)

and if A * B = D (another integer), then,

A * B/C = (A * B)/C = D/C

So if you have the number say (2/3), which you want to multiply by itself, then

(2/3) * (2/3) = (2*2)/(3*3) = 4/9

Thanks for your kind replies. I don't think you quite understood my point. I was talking about things like 2^3 being described like "Multiply 2 by itself three times". So, if we wanted to find the cubic root of 8, it would be 8^(1/3) which would be equivalent to saying "Multiply 8 by itself 0.3333 times". That's the part I didn't get. How can you multiply 8 by itself 0.3333 times?

 

[BigBeachBanana]

Can someone break this down for me, in layman terms and basic arithmetic?

Bisecting Method:
Say you want to approximate [imath]\sqrt{10}[/imath], and you know that [imath]\sqrt{10}\cdot \sqrt{10}=10[/imath]. Find a number times itself is close to 10.
Start the bisection between 0 and 10 i.e. 5, so 5*5=25 is too high.
Bisect again so 2.5*2.5= 6.25 is too low
Bisect again 3.75*3.75=14.0625 is still high
Continue the process until you get to your toleratable accuracy.

 

[JeffM]

I dislike intensely the explanation of powers as multiplying a number by itself n times.

Multiplication is a binary operation, which means it technically involves exactly two numbers. And [imath]3^2[/imath] means [imath]3 \times 3[/imath], which is ONE multiplication; there is no “multiply two times“ involved. And of course that ridiculous definition leaves students utterly confused when we talk about [imath]3^0[/imath]. How do you multiply something 0 times by itself. Here is a much better definition.

[math]\text {Given } a \text { is a real number} > 0 \text { and } n \text { is an integer} \ge 0, \text { then}\\ a^n = 1 \text { if } n = 0, \text { and }\\ a^n = a \times a^{(n-1)} \text { if } n > 0.[/math]
It is now possible and intuitive to extend that definition for exponentiation by any rational number, which is what I suspect you mean by decimal number.

EDIT: I shall leave to those with deeper faith than mine to explain what things like [imath]e^{\pi \times i}[/imath] mean.

 

[Apple30]

Bisecting Method:
Say you want to approximate [imath]\sqrt{10}[/imath], and you know that [imath]\sqrt{10}\cdot \sqrt{10}=10[/imath]. Find a number times itself is close to 10.
Start the bisection between 0 and 10 i.e. 5, so 5*5=25 is too high.
Bisect again so 2.5*2.5= 6.25 is too low
Bisect again 3.75*3.75=14.0625 is still high
Continue the process until you get to your toleratable accuracy.

Excellent explanation for how to do it manually, thanks very much!

Here is a much better definition.

[math]\text {Given } a \text { is a real number} > 0 \text { and } n \text { is an integer} \ge 0, \text { then}\\ a^n = 1 \text { if } n = 0, \text { and }\\ a^n = a \times a^{(n-1)} \text { if } n > 0.[/math]

Thanks very much for that definition, although I would point out that your definition is circular (you're trying to explain powers in a definition that depends on powers). It also isn't obvious why n^0 = 1. But perhaps I'm missing the point of definitions, which is just to show the rules.

I find it interesting how mathematicians write these definitions. As someone who is new to math, and studied programming, I always have to scratch my head when reading these definitions. Whenever mathematicians write things I feel like they are speaking backwards rather than forwards. For example, for the first line:

[math]a^n = 1 \text { if } n = 0[/math]

It would make much more sense to write it in English as "Any number to the power of zero will equal one", as opposed to "Any number to the power of something, equals one, if that something is zero". I wonder if this is because maths was invented by Italians or Greeks?

My programming brain is tempted to write it like:
a^(n=0) = 1
a^(n>0) = a^n

 

[Deleted member 4993]

I dislike intensely the explanation of powers as multiplying a number by itself n times.

Multiplication is a binary operation, which means it technically involves exactly two numbers. And [imath]3^2[/imath] means [imath]3 \times 3[/imath], which is ONE multiplication; there is no “multiply two times“ involved. And of course that ridiculous definition leaves students utterly confused when we talk about [imath]3^0[/imath]. How do you multiply something 0 times by itself. Here is a much better definition.

[math]\text {Given } a \text { is a real number} > 0 \text { and } n \text { is an integer} \ge 0, \text { then}\\ a^n = 1 \text { if } n = 0, \text { and }\\ a^n = a \times a^{(n-1)} \text { if } n > 0.[/math]
It is now possible and intuitive to extend that definition for exponentiation by any rational number, which is what I suspect you mean by decimal number.

EDIT: I shall leave to those with deeper faith than mine to explain what things like [imath]e^{\pi \times i}[/imath] mean.

I agree. "Multiplication" and "exponentiation" are two different operations.

Similarly, I dislike explaining multiplication as repeated addition. Then explaining "π * √2" will take some irrational effort.

 

[BigBeachBanana]

Similarly, I dislike explaining multiplication as repeated addition

Interested in your take on this. How would you introduce multiplication to someone, say your grandkids, who only know additions?

 

[Apple30]

Similarly, I dislike explaining multiplication as repeated addition. Then explaining "π * √2" will take some irrational effort.

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.

 

[JeffM]

I dislike intensely the explanation of powers as multiplying a number by itself n times.

Multiplication is a binary operation, which means it technically involves exactly two numbers. And [imath]3^2[/imath] means [imath]3 \times 3[/imath], which is ONE multiplication; there is no “multiply two times“ involved. And of course that ridiculous definition leaves students utterly confused when we talk about [imath]3^0[/imath]. How do you multiply something 0 times by itself. Here is a much better definition.

[math]\text {Given } a \text { is a real number} > 0 \text { and } n \text { is an integer} \ge 0, \text { then}\\ a^n = 1 \text { if } n = 0, \text { and }\\ a^n = a \times a^{(n-1)} \text { if } n > 0.[/math]
It is now possible and intuitive to extend that definition for exponentiation by any rational number, which is what I suspect you mean by decimal number.

Excellent explanation for how to do it manually, thanks very much!


Thanks very much for that definition, although I would point out that your definition is circular (you're trying to explain powers in a definition that depends on powers). It also isn't obvious why n^0 = 1. But perhaps I'm missing the point of definitions, which is just to show the rules.

I find it interesting how mathematicians write these definitions. As someone who is new to math, and studied programming, I always have to scratch my head when reading these definitions. Whenever mathematicians write things I feel like they are speaking backwards rather than forwards. For example, for the first line:

It would make much more sense to write it in English as "Any number to the power of zero will equal one", as opposed to "Any number to the power of something, equals one, if that something is zero". I wonder if this is because maths was invented by Italians or Greeks?

My programming brain is tempted to write it like:
a^(n=0) = 1
a^(n>0) = a^n

For now, I shall just point out that the definition is recursive rather than circular. To put it in programming terms, it is a terminating loop rather than an endless loop. Try programming it.

May have time later for a fuller response.

 

[Deleted member 4993]

Picard's algorithm:

guess √10 ~ 3 .......................... x_old

10/3 = 3.33333 ...................=x_calc

check ................ 3.33333 * 3.33333 = 11.111111 \(\displaystyle \ne\)10 ......................... Off...................

x_new = (x_old +x_new)/2 = 3.1666665 ......................... Off...................=x_calc

x_old = x_new

x_new = (x_old +x_new)/2 = 3.2499999 ......................... Off...................=x_calc ...........................and continue till x_calc's become close enough for you.

This algorithm is generally faster than bisection - however Picard works only for square root. For other roots use NewtonRaphson method. That process may wear out your calculator buttons or fingers. I still think

Easiest thing to do would be to buy a new calculator with a [x^y] button.

 

[Deleted member 4993]

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.

So please calculate π * √2 - using addition.

 

[Apple30]

For now, I shall just point out that the definition is recursive rather than circular. To put it in programming terms, it is a terminating loop rather than an endless loop. Try programming it.

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.

 

[Apple30]

So please calculate π * √2 - using addition.

3.14 * 1.41 = 3.14 + 1.28 = 4.4

 

[Deleted member 4993]

Interested in your take on this. How would you introduce multiplication to someone, say your grandkids, who only know additions?

The first introduction should be done through addition.

However, after they were introduced to irrational numbers - time comes to erase that explanation. Just like first introduction to "exponentiation" should be done through integers. At least that is how it was introduced to me and I introduced to my sons and grandsons (the second step involving irrational numbers has not come yet).

 

[BigBeachBanana]

The first introduction should be done through addition.

However, after they were introduced to irrational numbers - time comes to erase that explanation. Just like first introduction to "exponentiation" should be done through integers. At least that is how it was introduced to me and I introduced to my sons and grandsons (the second step involving irrational numbers has not come yet).

I think the OP was taught the same, but after s/he was introduced to irrational numbers. Nobody told s/he to erase that explanation. That led the OP to seek an explanation for

How to make sense of "Multiplying a number by itself decimal-number of times". What does that even mean?

 

[Deleted member 4993]

3.14 * 1.41 = 3.14 + 1.28 = 4.4

Incorrect

π * √2 ~ 4.442883

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

 

[Deleted member 4993]

(fractions are commonsense):

That is a news to me. For me, (3/4) is 3 operated by inverse of multiplication by 4.

π is an irrational number - cannot be expressed as fractions of integers. So is √2.