-0.000 012 668 315 321 211 57 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 012 668 315 321 211 57(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 57(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 57| = 0.000 012 668 315 321 211 57


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 57.

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 57 × 2 = 0 + 0.000 025 336 630 642 423 14;
  • 2) 0.000 025 336 630 642 423 14 × 2 = 0 + 0.000 050 673 261 284 846 28;
  • 3) 0.000 050 673 261 284 846 28 × 2 = 0 + 0.000 101 346 522 569 692 56;
  • 4) 0.000 101 346 522 569 692 56 × 2 = 0 + 0.000 202 693 045 139 385 12;
  • 5) 0.000 202 693 045 139 385 12 × 2 = 0 + 0.000 405 386 090 278 770 24;
  • 6) 0.000 405 386 090 278 770 24 × 2 = 0 + 0.000 810 772 180 557 540 48;
  • 7) 0.000 810 772 180 557 540 48 × 2 = 0 + 0.001 621 544 361 115 080 96;
  • 8) 0.001 621 544 361 115 080 96 × 2 = 0 + 0.003 243 088 722 230 161 92;
  • 9) 0.003 243 088 722 230 161 92 × 2 = 0 + 0.006 486 177 444 460 323 84;
  • 10) 0.006 486 177 444 460 323 84 × 2 = 0 + 0.012 972 354 888 920 647 68;
  • 11) 0.012 972 354 888 920 647 68 × 2 = 0 + 0.025 944 709 777 841 295 36;
  • 12) 0.025 944 709 777 841 295 36 × 2 = 0 + 0.051 889 419 555 682 590 72;
  • 13) 0.051 889 419 555 682 590 72 × 2 = 0 + 0.103 778 839 111 365 181 44;
  • 14) 0.103 778 839 111 365 181 44 × 2 = 0 + 0.207 557 678 222 730 362 88;
  • 15) 0.207 557 678 222 730 362 88 × 2 = 0 + 0.415 115 356 445 460 725 76;
  • 16) 0.415 115 356 445 460 725 76 × 2 = 0 + 0.830 230 712 890 921 451 52;
  • 17) 0.830 230 712 890 921 451 52 × 2 = 1 + 0.660 461 425 781 842 903 04;
  • 18) 0.660 461 425 781 842 903 04 × 2 = 1 + 0.320 922 851 563 685 806 08;
  • 19) 0.320 922 851 563 685 806 08 × 2 = 0 + 0.641 845 703 127 371 612 16;
  • 20) 0.641 845 703 127 371 612 16 × 2 = 1 + 0.283 691 406 254 743 224 32;
  • 21) 0.283 691 406 254 743 224 32 × 2 = 0 + 0.567 382 812 509 486 448 64;
  • 22) 0.567 382 812 509 486 448 64 × 2 = 1 + 0.134 765 625 018 972 897 28;
  • 23) 0.134 765 625 018 972 897 28 × 2 = 0 + 0.269 531 250 037 945 794 56;
  • 24) 0.269 531 250 037 945 794 56 × 2 = 0 + 0.539 062 500 075 891 589 12;
  • 25) 0.539 062 500 075 891 589 12 × 2 = 1 + 0.078 125 000 151 783 178 24;
  • 26) 0.078 125 000 151 783 178 24 × 2 = 0 + 0.156 250 000 303 566 356 48;
  • 27) 0.156 250 000 303 566 356 48 × 2 = 0 + 0.312 500 000 607 132 712 96;
  • 28) 0.312 500 000 607 132 712 96 × 2 = 0 + 0.625 000 001 214 265 425 92;
  • 29) 0.625 000 001 214 265 425 92 × 2 = 1 + 0.250 000 002 428 530 851 84;
  • 30) 0.250 000 002 428 530 851 84 × 2 = 0 + 0.500 000 004 857 061 703 68;
  • 31) 0.500 000 004 857 061 703 68 × 2 = 1 + 0.000 000 009 714 123 407 36;
  • 32) 0.000 000 009 714 123 407 36 × 2 = 0 + 0.000 000 019 428 246 814 72;
  • 33) 0.000 000 019 428 246 814 72 × 2 = 0 + 0.000 000 038 856 493 629 44;
  • 34) 0.000 000 038 856 493 629 44 × 2 = 0 + 0.000 000 077 712 987 258 88;
  • 35) 0.000 000 077 712 987 258 88 × 2 = 0 + 0.000 000 155 425 974 517 76;
  • 36) 0.000 000 155 425 974 517 76 × 2 = 0 + 0.000 000 310 851 949 035 52;
  • 37) 0.000 000 310 851 949 035 52 × 2 = 0 + 0.000 000 621 703 898 071 04;
  • 38) 0.000 000 621 703 898 071 04 × 2 = 0 + 0.000 001 243 407 796 142 08;
  • 39) 0.000 001 243 407 796 142 08 × 2 = 0 + 0.000 002 486 815 592 284 16;
  • 40) 0.000 002 486 815 592 284 16 × 2 = 0 + 0.000 004 973 631 184 568 32;

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 57(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 57(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 57(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 57 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