-0.000 000 000 742 147 673 37 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 37(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 37(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 37| = 0.000 000 000 742 147 673 37


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 37.

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 37 × 2 = 0 + 0.000 000 001 484 295 346 74;
  • 2) 0.000 000 001 484 295 346 74 × 2 = 0 + 0.000 000 002 968 590 693 48;
  • 3) 0.000 000 002 968 590 693 48 × 2 = 0 + 0.000 000 005 937 181 386 96;
  • 4) 0.000 000 005 937 181 386 96 × 2 = 0 + 0.000 000 011 874 362 773 92;
  • 5) 0.000 000 011 874 362 773 92 × 2 = 0 + 0.000 000 023 748 725 547 84;
  • 6) 0.000 000 023 748 725 547 84 × 2 = 0 + 0.000 000 047 497 451 095 68;
  • 7) 0.000 000 047 497 451 095 68 × 2 = 0 + 0.000 000 094 994 902 191 36;
  • 8) 0.000 000 094 994 902 191 36 × 2 = 0 + 0.000 000 189 989 804 382 72;
  • 9) 0.000 000 189 989 804 382 72 × 2 = 0 + 0.000 000 379 979 608 765 44;
  • 10) 0.000 000 379 979 608 765 44 × 2 = 0 + 0.000 000 759 959 217 530 88;
  • 11) 0.000 000 759 959 217 530 88 × 2 = 0 + 0.000 001 519 918 435 061 76;
  • 12) 0.000 001 519 918 435 061 76 × 2 = 0 + 0.000 003 039 836 870 123 52;
  • 13) 0.000 003 039 836 870 123 52 × 2 = 0 + 0.000 006 079 673 740 247 04;
  • 14) 0.000 006 079 673 740 247 04 × 2 = 0 + 0.000 012 159 347 480 494 08;
  • 15) 0.000 012 159 347 480 494 08 × 2 = 0 + 0.000 024 318 694 960 988 16;
  • 16) 0.000 024 318 694 960 988 16 × 2 = 0 + 0.000 048 637 389 921 976 32;
  • 17) 0.000 048 637 389 921 976 32 × 2 = 0 + 0.000 097 274 779 843 952 64;
  • 18) 0.000 097 274 779 843 952 64 × 2 = 0 + 0.000 194 549 559 687 905 28;
  • 19) 0.000 194 549 559 687 905 28 × 2 = 0 + 0.000 389 099 119 375 810 56;
  • 20) 0.000 389 099 119 375 810 56 × 2 = 0 + 0.000 778 198 238 751 621 12;
  • 21) 0.000 778 198 238 751 621 12 × 2 = 0 + 0.001 556 396 477 503 242 24;
  • 22) 0.001 556 396 477 503 242 24 × 2 = 0 + 0.003 112 792 955 006 484 48;
  • 23) 0.003 112 792 955 006 484 48 × 2 = 0 + 0.006 225 585 910 012 968 96;
  • 24) 0.006 225 585 910 012 968 96 × 2 = 0 + 0.012 451 171 820 025 937 92;
  • 25) 0.012 451 171 820 025 937 92 × 2 = 0 + 0.024 902 343 640 051 875 84;
  • 26) 0.024 902 343 640 051 875 84 × 2 = 0 + 0.049 804 687 280 103 751 68;
  • 27) 0.049 804 687 280 103 751 68 × 2 = 0 + 0.099 609 374 560 207 503 36;
  • 28) 0.099 609 374 560 207 503 36 × 2 = 0 + 0.199 218 749 120 415 006 72;
  • 29) 0.199 218 749 120 415 006 72 × 2 = 0 + 0.398 437 498 240 830 013 44;
  • 30) 0.398 437 498 240 830 013 44 × 2 = 0 + 0.796 874 996 481 660 026 88;
  • 31) 0.796 874 996 481 660 026 88 × 2 = 1 + 0.593 749 992 963 320 053 76;
  • 32) 0.593 749 992 963 320 053 76 × 2 = 1 + 0.187 499 985 926 640 107 52;
  • 33) 0.187 499 985 926 640 107 52 × 2 = 0 + 0.374 999 971 853 280 215 04;
  • 34) 0.374 999 971 853 280 215 04 × 2 = 0 + 0.749 999 943 706 560 430 08;
  • 35) 0.749 999 943 706 560 430 08 × 2 = 1 + 0.499 999 887 413 120 860 16;
  • 36) 0.499 999 887 413 120 860 16 × 2 = 0 + 0.999 999 774 826 241 720 32;
  • 37) 0.999 999 774 826 241 720 32 × 2 = 1 + 0.999 999 549 652 483 440 64;
  • 38) 0.999 999 549 652 483 440 64 × 2 = 1 + 0.999 999 099 304 966 881 28;
  • 39) 0.999 999 099 304 966 881 28 × 2 = 1 + 0.999 998 198 609 933 762 56;
  • 40) 0.999 998 198 609 933 762 56 × 2 = 1 + 0.999 996 397 219 867 525 12;
  • 41) 0.999 996 397 219 867 525 12 × 2 = 1 + 0.999 992 794 439 735 050 24;
  • 42) 0.999 992 794 439 735 050 24 × 2 = 1 + 0.999 985 588 879 470 100 48;
  • 43) 0.999 985 588 879 470 100 48 × 2 = 1 + 0.999 971 177 758 940 200 96;
  • 44) 0.999 971 177 758 940 200 96 × 2 = 1 + 0.999 942 355 517 880 401 92;
  • 45) 0.999 942 355 517 880 401 92 × 2 = 1 + 0.999 884 711 035 760 803 84;
  • 46) 0.999 884 711 035 760 803 84 × 2 = 1 + 0.999 769 422 071 521 607 68;
  • 47) 0.999 769 422 071 521 607 68 × 2 = 1 + 0.999 538 844 143 043 215 36;
  • 48) 0.999 538 844 143 043 215 36 × 2 = 1 + 0.999 077 688 286 086 430 72;
  • 49) 0.999 077 688 286 086 430 72 × 2 = 1 + 0.998 155 376 572 172 861 44;
  • 50) 0.998 155 376 572 172 861 44 × 2 = 1 + 0.996 310 753 144 345 722 88;
  • 51) 0.996 310 753 144 345 722 88 × 2 = 1 + 0.992 621 506 288 691 445 76;
  • 52) 0.992 621 506 288 691 445 76 × 2 = 1 + 0.985 243 012 577 382 891 52;
  • 53) 0.985 243 012 577 382 891 52 × 2 = 1 + 0.970 486 025 154 765 783 04;
  • 54) 0.970 486 025 154 765 783 04 × 2 = 1 + 0.940 972 050 309 531 566 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 37(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 37(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 37(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 37 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