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Prove that you can multiply using these steps

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1

$begingroup$

How to prove this?

1. Think of a simple multiplication problem, I will use $22 times 7$




2. Divide the first factor by 2; if the number isn't a whole number, floor it (e. g. $11 / 2 = 5$), until you come to $1$




3. Multiply the second factor by $2$ with increasing power ($7 times 2^0, 7 times 2^1$ etc.) until you make as many numbers as you made in step 1, make "columns".




4. Remove every column that has an even number in its upper cell.




5. Add all number in the second line together.

This works universally for whole numbers.

e. g.

$22;11;5;2;1$

$7;14;28;56;112$

this becomes

$11;5;1$

$14;28;112$

$22 times 7 = 14 + 28 + 112 = 154$

Thanks in advance!

arithmetic

share|cite|improve this question

edited Mar 21 at 14:29

J. W. Tanner

4,4711320

asked Mar 21 at 12:59

Xxx DddXxx Ddd

454

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This has to do with binary multiplication. See for example, this answer of mine : math.stackexchange.com/questions/3122152/… especially the latter half on the Vakil method
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Observe $22=2+4+16$ and $7(2+4+16)=14+28+112$
  $endgroup$
  – Daniel Mathias
  Mar 21 at 13:40
  

  
  
  
  

add a comment | 

1

$begingroup$

How to prove this?

1. Think of a simple multiplication problem, I will use $22 times 7$




2. Divide the first factor by 2; if the number isn't a whole number, floor it (e. g. $11 / 2 = 5$), until you come to $1$




3. Multiply the second factor by $2$ with increasing power ($7 times 2^0, 7 times 2^1$ etc.) until you make as many numbers as you made in step 1, make "columns".




4. Remove every column that has an even number in its upper cell.




5. Add all number in the second line together.

This works universally for whole numbers.

e. g.

$22;11;5;2;1$

$7;14;28;56;112$

this becomes

$11;5;1$

$14;28;112$

$22 times 7 = 14 + 28 + 112 = 154$

Thanks in advance!

arithmetic

share|cite|improve this question

edited Mar 21 at 14:29

J. W. Tanner

4,4711320

asked Mar 21 at 12:59

Xxx DddXxx Ddd

454

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This has to do with binary multiplication. See for example, this answer of mine : math.stackexchange.com/questions/3122152/… especially the latter half on the Vakil method
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Observe $22=2+4+16$ and $7(2+4+16)=14+28+112$
  $endgroup$
  – Daniel Mathias
  Mar 21 at 13:40
  

  
  
  
  

add a comment | 

$begingroup$

How to prove this?

1. Think of a simple multiplication problem, I will use $22 times 7$




2. Divide the first factor by 2; if the number isn't a whole number, floor it (e. g. $11 / 2 = 5$), until you come to $1$




3. Multiply the second factor by $2$ with increasing power ($7 times 2^0, 7 times 2^1$ etc.) until you make as many numbers as you made in step 1, make "columns".




4. Remove every column that has an even number in its upper cell.




5. Add all number in the second line together.

This works universally for whole numbers.

e. g.

$22;11;5;2;1$

$7;14;28;56;112$

this becomes

$11;5;1$

$14;28;112$

$22 times 7 = 14 + 28 + 112 = 154$

Thanks in advance!

arithmetic

share|cite|improve this question

edited Mar 21 at 14:29

J. W. Tanner

4,4711320

asked Mar 21 at 12:59

Xxx DddXxx Ddd

454

$endgroup$

share|cite|improve this question

edited Mar 21 at 14:29

J. W. Tanner

4,4711320

asked Mar 21 at 12:59

Xxx DddXxx Ddd

454

share|cite|improve this question

edited Mar 21 at 14:29

J. W. Tanner

4,4711320

asked Mar 21 at 12:59

Xxx DddXxx Ddd

454

edited Mar 21 at 14:29

J. W. Tanner

4,4711320

edited Mar 21 at 14:29

J. W. Tanner

4,4711320

asked Mar 21 at 12:59

Xxx DddXxx Ddd

454

asked Mar 21 at 12:59

Xxx DddXxx Ddd

454