-0.000 000 000 742 147 673 87 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 87(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
-0.000 000 000 742 147 673 87(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:

|-0.000 000 000 742 147 673 87| = 0.000 000 000 742 147 673 87


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: 0.000 000 000 742 147 673 87.

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) 0.000 000 000 742 147 673 87 × 2 = 0 + 0.000 000 001 484 295 347 74;
  • 2) 0.000 000 001 484 295 347 74 × 2 = 0 + 0.000 000 002 968 590 695 48;
  • 3) 0.000 000 002 968 590 695 48 × 2 = 0 + 0.000 000 005 937 181 390 96;
  • 4) 0.000 000 005 937 181 390 96 × 2 = 0 + 0.000 000 011 874 362 781 92;
  • 5) 0.000 000 011 874 362 781 92 × 2 = 0 + 0.000 000 023 748 725 563 84;
  • 6) 0.000 000 023 748 725 563 84 × 2 = 0 + 0.000 000 047 497 451 127 68;
  • 7) 0.000 000 047 497 451 127 68 × 2 = 0 + 0.000 000 094 994 902 255 36;
  • 8) 0.000 000 094 994 902 255 36 × 2 = 0 + 0.000 000 189 989 804 510 72;
  • 9) 0.000 000 189 989 804 510 72 × 2 = 0 + 0.000 000 379 979 609 021 44;
  • 10) 0.000 000 379 979 609 021 44 × 2 = 0 + 0.000 000 759 959 218 042 88;
  • 11) 0.000 000 759 959 218 042 88 × 2 = 0 + 0.000 001 519 918 436 085 76;
  • 12) 0.000 001 519 918 436 085 76 × 2 = 0 + 0.000 003 039 836 872 171 52;
  • 13) 0.000 003 039 836 872 171 52 × 2 = 0 + 0.000 006 079 673 744 343 04;
  • 14) 0.000 006 079 673 744 343 04 × 2 = 0 + 0.000 012 159 347 488 686 08;
  • 15) 0.000 012 159 347 488 686 08 × 2 = 0 + 0.000 024 318 694 977 372 16;
  • 16) 0.000 024 318 694 977 372 16 × 2 = 0 + 0.000 048 637 389 954 744 32;
  • 17) 0.000 048 637 389 954 744 32 × 2 = 0 + 0.000 097 274 779 909 488 64;
  • 18) 0.000 097 274 779 909 488 64 × 2 = 0 + 0.000 194 549 559 818 977 28;
  • 19) 0.000 194 549 559 818 977 28 × 2 = 0 + 0.000 389 099 119 637 954 56;
  • 20) 0.000 389 099 119 637 954 56 × 2 = 0 + 0.000 778 198 239 275 909 12;
  • 21) 0.000 778 198 239 275 909 12 × 2 = 0 + 0.001 556 396 478 551 818 24;
  • 22) 0.001 556 396 478 551 818 24 × 2 = 0 + 0.003 112 792 957 103 636 48;
  • 23) 0.003 112 792 957 103 636 48 × 2 = 0 + 0.006 225 585 914 207 272 96;
  • 24) 0.006 225 585 914 207 272 96 × 2 = 0 + 0.012 451 171 828 414 545 92;
  • 25) 0.012 451 171 828 414 545 92 × 2 = 0 + 0.024 902 343 656 829 091 84;
  • 26) 0.024 902 343 656 829 091 84 × 2 = 0 + 0.049 804 687 313 658 183 68;
  • 27) 0.049 804 687 313 658 183 68 × 2 = 0 + 0.099 609 374 627 316 367 36;
  • 28) 0.099 609 374 627 316 367 36 × 2 = 0 + 0.199 218 749 254 632 734 72;
  • 29) 0.199 218 749 254 632 734 72 × 2 = 0 + 0.398 437 498 509 265 469 44;
  • 30) 0.398 437 498 509 265 469 44 × 2 = 0 + 0.796 874 997 018 530 938 88;
  • 31) 0.796 874 997 018 530 938 88 × 2 = 1 + 0.593 749 994 037 061 877 76;
  • 32) 0.593 749 994 037 061 877 76 × 2 = 1 + 0.187 499 988 074 123 755 52;
  • 33) 0.187 499 988 074 123 755 52 × 2 = 0 + 0.374 999 976 148 247 511 04;
  • 34) 0.374 999 976 148 247 511 04 × 2 = 0 + 0.749 999 952 296 495 022 08;
  • 35) 0.749 999 952 296 495 022 08 × 2 = 1 + 0.499 999 904 592 990 044 16;
  • 36) 0.499 999 904 592 990 044 16 × 2 = 0 + 0.999 999 809 185 980 088 32;
  • 37) 0.999 999 809 185 980 088 32 × 2 = 1 + 0.999 999 618 371 960 176 64;
  • 38) 0.999 999 618 371 960 176 64 × 2 = 1 + 0.999 999 236 743 920 353 28;
  • 39) 0.999 999 236 743 920 353 28 × 2 = 1 + 0.999 998 473 487 840 706 56;
  • 40) 0.999 998 473 487 840 706 56 × 2 = 1 + 0.999 996 946 975 681 413 12;
  • 41) 0.999 996 946 975 681 413 12 × 2 = 1 + 0.999 993 893 951 362 826 24;
  • 42) 0.999 993 893 951 362 826 24 × 2 = 1 + 0.999 987 787 902 725 652 48;
  • 43) 0.999 987 787 902 725 652 48 × 2 = 1 + 0.999 975 575 805 451 304 96;
  • 44) 0.999 975 575 805 451 304 96 × 2 = 1 + 0.999 951 151 610 902 609 92;
  • 45) 0.999 951 151 610 902 609 92 × 2 = 1 + 0.999 902 303 221 805 219 84;
  • 46) 0.999 902 303 221 805 219 84 × 2 = 1 + 0.999 804 606 443 610 439 68;
  • 47) 0.999 804 606 443 610 439 68 × 2 = 1 + 0.999 609 212 887 220 879 36;
  • 48) 0.999 609 212 887 220 879 36 × 2 = 1 + 0.999 218 425 774 441 758 72;
  • 49) 0.999 218 425 774 441 758 72 × 2 = 1 + 0.998 436 851 548 883 517 44;
  • 50) 0.998 436 851 548 883 517 44 × 2 = 1 + 0.996 873 703 097 767 034 88;
  • 51) 0.996 873 703 097 767 034 88 × 2 = 1 + 0.993 747 406 195 534 069 76;
  • 52) 0.993 747 406 195 534 069 76 × 2 = 1 + 0.987 494 812 391 068 139 52;
  • 53) 0.987 494 812 391 068 139 52 × 2 = 1 + 0.974 989 624 782 136 279 04;
  • 54) 0.974 989 624 782 136 279 04 × 2 = 1 + 0.949 979 249 564 272 558 08;

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:


0.000 000 000 742 147 673 87(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 673 87(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 673 87(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 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 0111 1111 1111 1111 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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 673 87 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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