Respuesta :

Respuesta:

Para convertir números del sistema binario al sistema decimal, se debe usar el concepto de valor posicional. Cada dígito binario representa una potencia de 2, comenzando desde la derecha con la potencia de 2 elevada a la potencia de 0 (1), luego la potencia de 2 elevada a la potencia de 1 (2), y así sucesivamente.

10

1

2

101

2

:

10

1

2

=

(

1

×

2

2

)

+

(

0

×

2

1

)

+

(

1

×

2

0

)

101

2

=(1×2

2

)+(0×2

1

)+(1×2

0

)

=

(

1

×

4

)

+

(

0

×

2

)

+

(

1

×

1

)

=(1×4)+(0×2)+(1×1)

=

4

+

0

+

1

=4+0+1

=

5

10

=5

10

110

1

2

1101

2

:

110

1

2

=

(

1

×

2

3

)

+

(

1

×

2

2

)

+

(

0

×

2

1

)

+

(

1

×

2

0

)

1101

2

=(1×2

3

)+(1×2

2

)+(0×2

1

)+(1×2

0

)

=

(

1

×

8

)

+

(

1

×

4

)

+

(

0

×

2

)

+

(

1

×

1

)

=(1×8)+(1×4)+(0×2)+(1×1)

=

8

+

4

+

0

+

1

=8+4+0+1

=

1

3

10

=13

10

1000

1

2

10001

2

:

1000

1

2

=

(

1

×

2

4

)

+

(

0

×

2

3

)

+

(

0

×

2

2

)

+

(

0

×

2

1

)

+

(

1

×

2

0

)

10001

2

=(1×2

4

)+(0×2

3

)+(0×2

2

)+(0×2

1

)+(1×2

0

)

=

(

1

×

16

)

+

(

0

×

8

)

+

(

0

×

4

)

+

(

0

×

2

)

+

(

1

×

1

)

=(1×16)+(0×8)+(0×4)+(0×2)+(1×1)

=

16

+

0

+

0

+

0

+

1

=16+0+0+0+1

=

1

7

10

=17

10

11110

1

2

111101

2

:

11110

1

2

=

(

1

×

2

5

)

+

(

1

×

2

4

)

+

(

1

×

2

3

)

+

(

1

×

2

2

)

+

(

0

×

2

1

)

+

(

1

×

2

0

)

111101

2

=(1×2

5

)+(1×2

4

)+(1×2

3

)+(1×2

2

)+(0×2

1

)+(1×2

0

)

=

(

1

×

32

)

+

(

1

×

16

)

+

(

1

×

8

)

+

(

1

×

4

)

+

(

0

×

2

)

+

(

1

×

1

)

=(1×32)+(1×16)+(1×8)+(1×4)+(0×2)+(1×1)

=

32

+

16

+

8

+

4

+

0

+

1

=32+16+8+4+0+1

=

6

1

10

=61

10

100000000

1

2

1000000001

2

:

100000000

1

2

=

(

1

×

2

9

)

+

(

0

×

2

8

)

+

(

0

×

2

7

)

+

(

0

×

2

6

)

+

(

0

×

2

5

)

+

(

0

×

2

4

)

+

(

0

×

2

3

)

+

(

0

×

2

2

)

+

(

0

×

2

1

)

+

(

1

×

2

0

)

1000000001

2

=(1×2

9

)+(0×2

8

)+(0×2

7

)+(0×2

6

)+(0×2

5

)+(0×2

4

)+(0×2

3

)+(0×2

2

)+(0×2

1

)+(1×2

0

)

=

(

1

×

512

)

+

(

0

×

256

)

+

(

0

×

128

)

+

(

0

×

64

)

+

(

0

×

32

)

+

(

0

×

16

)

+

(

0

×

8

)

+

(

0

×

4

)

+

(

0

×

2

)

+

(

1

×

1

)

=(1×512)+(0×256)+(0×128)+(0×64)+(0×32)+(0×16)+(0×8)+(0×4)+(0×2)+(1×1)

=

512

+

0

+

0

+

0

+

0

+

0

+

0

+

0

+

0

+

1

=512+0+0+0+0+0+0+0+0+1

=

51

3

10

=513

10

Por lo tanto, los números binarios dados en su forma decimal son:

10

1

2

=

5

10

101

2

=5

10

110

1

2

=

1

3

10

1101

2

=13

10

1000

1

2

=

1

7

10

10001

2

=17

10

11110

1

2

=

6

1

10

111101

2

=61

10

100000000

1

2

=

51

3

10

1000000001

2

=513

10

Explicación paso a paso: