-0.000 012 668 315 321 211 19 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 012 668 315 321 211 19(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 012 668 315 321 211 19(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 012 668 315 321 211 19| = 0.000 012 668 315 321 211 19


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 012 668 315 321 211 19.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 012 668 315 321 211 19 × 2 = 0 + 0.000 025 336 630 642 422 38;
  • 2) 0.000 025 336 630 642 422 38 × 2 = 0 + 0.000 050 673 261 284 844 76;
  • 3) 0.000 050 673 261 284 844 76 × 2 = 0 + 0.000 101 346 522 569 689 52;
  • 4) 0.000 101 346 522 569 689 52 × 2 = 0 + 0.000 202 693 045 139 379 04;
  • 5) 0.000 202 693 045 139 379 04 × 2 = 0 + 0.000 405 386 090 278 758 08;
  • 6) 0.000 405 386 090 278 758 08 × 2 = 0 + 0.000 810 772 180 557 516 16;
  • 7) 0.000 810 772 180 557 516 16 × 2 = 0 + 0.001 621 544 361 115 032 32;
  • 8) 0.001 621 544 361 115 032 32 × 2 = 0 + 0.003 243 088 722 230 064 64;
  • 9) 0.003 243 088 722 230 064 64 × 2 = 0 + 0.006 486 177 444 460 129 28;
  • 10) 0.006 486 177 444 460 129 28 × 2 = 0 + 0.012 972 354 888 920 258 56;
  • 11) 0.012 972 354 888 920 258 56 × 2 = 0 + 0.025 944 709 777 840 517 12;
  • 12) 0.025 944 709 777 840 517 12 × 2 = 0 + 0.051 889 419 555 681 034 24;
  • 13) 0.051 889 419 555 681 034 24 × 2 = 0 + 0.103 778 839 111 362 068 48;
  • 14) 0.103 778 839 111 362 068 48 × 2 = 0 + 0.207 557 678 222 724 136 96;
  • 15) 0.207 557 678 222 724 136 96 × 2 = 0 + 0.415 115 356 445 448 273 92;
  • 16) 0.415 115 356 445 448 273 92 × 2 = 0 + 0.830 230 712 890 896 547 84;
  • 17) 0.830 230 712 890 896 547 84 × 2 = 1 + 0.660 461 425 781 793 095 68;
  • 18) 0.660 461 425 781 793 095 68 × 2 = 1 + 0.320 922 851 563 586 191 36;
  • 19) 0.320 922 851 563 586 191 36 × 2 = 0 + 0.641 845 703 127 172 382 72;
  • 20) 0.641 845 703 127 172 382 72 × 2 = 1 + 0.283 691 406 254 344 765 44;
  • 21) 0.283 691 406 254 344 765 44 × 2 = 0 + 0.567 382 812 508 689 530 88;
  • 22) 0.567 382 812 508 689 530 88 × 2 = 1 + 0.134 765 625 017 379 061 76;
  • 23) 0.134 765 625 017 379 061 76 × 2 = 0 + 0.269 531 250 034 758 123 52;
  • 24) 0.269 531 250 034 758 123 52 × 2 = 0 + 0.539 062 500 069 516 247 04;
  • 25) 0.539 062 500 069 516 247 04 × 2 = 1 + 0.078 125 000 139 032 494 08;
  • 26) 0.078 125 000 139 032 494 08 × 2 = 0 + 0.156 250 000 278 064 988 16;
  • 27) 0.156 250 000 278 064 988 16 × 2 = 0 + 0.312 500 000 556 129 976 32;
  • 28) 0.312 500 000 556 129 976 32 × 2 = 0 + 0.625 000 001 112 259 952 64;
  • 29) 0.625 000 001 112 259 952 64 × 2 = 1 + 0.250 000 002 224 519 905 28;
  • 30) 0.250 000 002 224 519 905 28 × 2 = 0 + 0.500 000 004 449 039 810 56;
  • 31) 0.500 000 004 449 039 810 56 × 2 = 1 + 0.000 000 008 898 079 621 12;
  • 32) 0.000 000 008 898 079 621 12 × 2 = 0 + 0.000 000 017 796 159 242 24;
  • 33) 0.000 000 017 796 159 242 24 × 2 = 0 + 0.000 000 035 592 318 484 48;
  • 34) 0.000 000 035 592 318 484 48 × 2 = 0 + 0.000 000 071 184 636 968 96;
  • 35) 0.000 000 071 184 636 968 96 × 2 = 0 + 0.000 000 142 369 273 937 92;
  • 36) 0.000 000 142 369 273 937 92 × 2 = 0 + 0.000 000 284 738 547 875 84;
  • 37) 0.000 000 284 738 547 875 84 × 2 = 0 + 0.000 000 569 477 095 751 68;
  • 38) 0.000 000 569 477 095 751 68 × 2 = 0 + 0.000 001 138 954 191 503 36;
  • 39) 0.000 001 138 954 191 503 36 × 2 = 0 + 0.000 002 277 908 383 006 72;
  • 40) 0.000 002 277 908 383 006 72 × 2 = 0 + 0.000 004 555 816 766 013 44;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 012 668 315 321 211 19(10) =

0.0000 0000 0000 0000 1101 0100 1000 1010 0000 0000(2)

6. Positive number before normalization:

0.000 012 668 315 321 211 19(10) =

0.0000 0000 0000 0000 1101 0100 1000 1010 0000 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 012 668 315 321 211 19(10) =

0.0000 0000 0000 0000 1101 0100 1000 1010 0000 0000(2) =

0.0000 0000 0000 0000 1101 0100 1000 1010 0000 0000(2) × 20 =

1.1010 1001 0001 0100 0000 000(2) × 2-17


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -17

Mantissa (not normalized):
1.1010 1001 0001 0100 0000 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-17 + 2(8-1) - 1 =

(-17 + 127)(10) =

110(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 110 ÷ 2 = 55 + 0;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

110(10) =

0110 1110(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 101 0100 1000 1010 0000 0000 =

101 0100 1000 1010 0000 0000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 1110

Mantissa (23 bits) =
101 0100 1000 1010 0000 0000


Decimal number -0.000 012 668 315 321 211 19 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1110 - 101 0100 1000 1010 0000 0000

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111