-0.000 000 000 742 148 07 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 07(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
-0.000 000 000 742 148 07(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:

|-0.000 000 000 742 148 07| = 0.000 000 000 742 148 07


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: 0.000 000 000 742 148 07.

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) 0.000 000 000 742 148 07 × 2 = 0 + 0.000 000 001 484 296 14;
  • 2) 0.000 000 001 484 296 14 × 2 = 0 + 0.000 000 002 968 592 28;
  • 3) 0.000 000 002 968 592 28 × 2 = 0 + 0.000 000 005 937 184 56;
  • 4) 0.000 000 005 937 184 56 × 2 = 0 + 0.000 000 011 874 369 12;
  • 5) 0.000 000 011 874 369 12 × 2 = 0 + 0.000 000 023 748 738 24;
  • 6) 0.000 000 023 748 738 24 × 2 = 0 + 0.000 000 047 497 476 48;
  • 7) 0.000 000 047 497 476 48 × 2 = 0 + 0.000 000 094 994 952 96;
  • 8) 0.000 000 094 994 952 96 × 2 = 0 + 0.000 000 189 989 905 92;
  • 9) 0.000 000 189 989 905 92 × 2 = 0 + 0.000 000 379 979 811 84;
  • 10) 0.000 000 379 979 811 84 × 2 = 0 + 0.000 000 759 959 623 68;
  • 11) 0.000 000 759 959 623 68 × 2 = 0 + 0.000 001 519 919 247 36;
  • 12) 0.000 001 519 919 247 36 × 2 = 0 + 0.000 003 039 838 494 72;
  • 13) 0.000 003 039 838 494 72 × 2 = 0 + 0.000 006 079 676 989 44;
  • 14) 0.000 006 079 676 989 44 × 2 = 0 + 0.000 012 159 353 978 88;
  • 15) 0.000 012 159 353 978 88 × 2 = 0 + 0.000 024 318 707 957 76;
  • 16) 0.000 024 318 707 957 76 × 2 = 0 + 0.000 048 637 415 915 52;
  • 17) 0.000 048 637 415 915 52 × 2 = 0 + 0.000 097 274 831 831 04;
  • 18) 0.000 097 274 831 831 04 × 2 = 0 + 0.000 194 549 663 662 08;
  • 19) 0.000 194 549 663 662 08 × 2 = 0 + 0.000 389 099 327 324 16;
  • 20) 0.000 389 099 327 324 16 × 2 = 0 + 0.000 778 198 654 648 32;
  • 21) 0.000 778 198 654 648 32 × 2 = 0 + 0.001 556 397 309 296 64;
  • 22) 0.001 556 397 309 296 64 × 2 = 0 + 0.003 112 794 618 593 28;
  • 23) 0.003 112 794 618 593 28 × 2 = 0 + 0.006 225 589 237 186 56;
  • 24) 0.006 225 589 237 186 56 × 2 = 0 + 0.012 451 178 474 373 12;
  • 25) 0.012 451 178 474 373 12 × 2 = 0 + 0.024 902 356 948 746 24;
  • 26) 0.024 902 356 948 746 24 × 2 = 0 + 0.049 804 713 897 492 48;
  • 27) 0.049 804 713 897 492 48 × 2 = 0 + 0.099 609 427 794 984 96;
  • 28) 0.099 609 427 794 984 96 × 2 = 0 + 0.199 218 855 589 969 92;
  • 29) 0.199 218 855 589 969 92 × 2 = 0 + 0.398 437 711 179 939 84;
  • 30) 0.398 437 711 179 939 84 × 2 = 0 + 0.796 875 422 359 879 68;
  • 31) 0.796 875 422 359 879 68 × 2 = 1 + 0.593 750 844 719 759 36;
  • 32) 0.593 750 844 719 759 36 × 2 = 1 + 0.187 501 689 439 518 72;
  • 33) 0.187 501 689 439 518 72 × 2 = 0 + 0.375 003 378 879 037 44;
  • 34) 0.375 003 378 879 037 44 × 2 = 0 + 0.750 006 757 758 074 88;
  • 35) 0.750 006 757 758 074 88 × 2 = 1 + 0.500 013 515 516 149 76;
  • 36) 0.500 013 515 516 149 76 × 2 = 1 + 0.000 027 031 032 299 52;
  • 37) 0.000 027 031 032 299 52 × 2 = 0 + 0.000 054 062 064 599 04;
  • 38) 0.000 054 062 064 599 04 × 2 = 0 + 0.000 108 124 129 198 08;
  • 39) 0.000 108 124 129 198 08 × 2 = 0 + 0.000 216 248 258 396 16;
  • 40) 0.000 216 248 258 396 16 × 2 = 0 + 0.000 432 496 516 792 32;
  • 41) 0.000 432 496 516 792 32 × 2 = 0 + 0.000 864 993 033 584 64;
  • 42) 0.000 864 993 033 584 64 × 2 = 0 + 0.001 729 986 067 169 28;
  • 43) 0.001 729 986 067 169 28 × 2 = 0 + 0.003 459 972 134 338 56;
  • 44) 0.003 459 972 134 338 56 × 2 = 0 + 0.006 919 944 268 677 12;
  • 45) 0.006 919 944 268 677 12 × 2 = 0 + 0.013 839 888 537 354 24;
  • 46) 0.013 839 888 537 354 24 × 2 = 0 + 0.027 679 777 074 708 48;
  • 47) 0.027 679 777 074 708 48 × 2 = 0 + 0.055 359 554 149 416 96;
  • 48) 0.055 359 554 149 416 96 × 2 = 0 + 0.110 719 108 298 833 92;
  • 49) 0.110 719 108 298 833 92 × 2 = 0 + 0.221 438 216 597 667 84;
  • 50) 0.221 438 216 597 667 84 × 2 = 0 + 0.442 876 433 195 335 68;
  • 51) 0.442 876 433 195 335 68 × 2 = 0 + 0.885 752 866 390 671 36;
  • 52) 0.885 752 866 390 671 36 × 2 = 1 + 0.771 505 732 781 342 72;
  • 53) 0.771 505 732 781 342 72 × 2 = 1 + 0.543 011 465 562 685 44;
  • 54) 0.543 011 465 562 685 44 × 2 = 1 + 0.086 022 931 125 370 88;

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:


0.000 000 000 742 148 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 11(2)

6. Positive number before normalization:

0.000 000 000 742 148 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 148 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 11(2) × 20 =

1.1001 1000 0000 0000 0000 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 1000 0000 0000 0000 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

96(10) =

0110 0000(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. 100 1100 0000 0000 0000 0111 =

100 1100 0000 0000 0000 0111


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 0000

Mantissa (23 bits) =
100 1100 0000 0000 0000 0111


Decimal number -0.000 000 000 742 148 07 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0111

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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