-0.000 000 000 742 147 685 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 685 6(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
-0.000 000 000 742 147 685 6(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:

|-0.000 000 000 742 147 685 6| = 0.000 000 000 742 147 685 6


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: 0.000 000 000 742 147 685 6.

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) 0.000 000 000 742 147 685 6 × 2 = 0 + 0.000 000 001 484 295 371 2;
  • 2) 0.000 000 001 484 295 371 2 × 2 = 0 + 0.000 000 002 968 590 742 4;
  • 3) 0.000 000 002 968 590 742 4 × 2 = 0 + 0.000 000 005 937 181 484 8;
  • 4) 0.000 000 005 937 181 484 8 × 2 = 0 + 0.000 000 011 874 362 969 6;
  • 5) 0.000 000 011 874 362 969 6 × 2 = 0 + 0.000 000 023 748 725 939 2;
  • 6) 0.000 000 023 748 725 939 2 × 2 = 0 + 0.000 000 047 497 451 878 4;
  • 7) 0.000 000 047 497 451 878 4 × 2 = 0 + 0.000 000 094 994 903 756 8;
  • 8) 0.000 000 094 994 903 756 8 × 2 = 0 + 0.000 000 189 989 807 513 6;
  • 9) 0.000 000 189 989 807 513 6 × 2 = 0 + 0.000 000 379 979 615 027 2;
  • 10) 0.000 000 379 979 615 027 2 × 2 = 0 + 0.000 000 759 959 230 054 4;
  • 11) 0.000 000 759 959 230 054 4 × 2 = 0 + 0.000 001 519 918 460 108 8;
  • 12) 0.000 001 519 918 460 108 8 × 2 = 0 + 0.000 003 039 836 920 217 6;
  • 13) 0.000 003 039 836 920 217 6 × 2 = 0 + 0.000 006 079 673 840 435 2;
  • 14) 0.000 006 079 673 840 435 2 × 2 = 0 + 0.000 012 159 347 680 870 4;
  • 15) 0.000 012 159 347 680 870 4 × 2 = 0 + 0.000 024 318 695 361 740 8;
  • 16) 0.000 024 318 695 361 740 8 × 2 = 0 + 0.000 048 637 390 723 481 6;
  • 17) 0.000 048 637 390 723 481 6 × 2 = 0 + 0.000 097 274 781 446 963 2;
  • 18) 0.000 097 274 781 446 963 2 × 2 = 0 + 0.000 194 549 562 893 926 4;
  • 19) 0.000 194 549 562 893 926 4 × 2 = 0 + 0.000 389 099 125 787 852 8;
  • 20) 0.000 389 099 125 787 852 8 × 2 = 0 + 0.000 778 198 251 575 705 6;
  • 21) 0.000 778 198 251 575 705 6 × 2 = 0 + 0.001 556 396 503 151 411 2;
  • 22) 0.001 556 396 503 151 411 2 × 2 = 0 + 0.003 112 793 006 302 822 4;
  • 23) 0.003 112 793 006 302 822 4 × 2 = 0 + 0.006 225 586 012 605 644 8;
  • 24) 0.006 225 586 012 605 644 8 × 2 = 0 + 0.012 451 172 025 211 289 6;
  • 25) 0.012 451 172 025 211 289 6 × 2 = 0 + 0.024 902 344 050 422 579 2;
  • 26) 0.024 902 344 050 422 579 2 × 2 = 0 + 0.049 804 688 100 845 158 4;
  • 27) 0.049 804 688 100 845 158 4 × 2 = 0 + 0.099 609 376 201 690 316 8;
  • 28) 0.099 609 376 201 690 316 8 × 2 = 0 + 0.199 218 752 403 380 633 6;
  • 29) 0.199 218 752 403 380 633 6 × 2 = 0 + 0.398 437 504 806 761 267 2;
  • 30) 0.398 437 504 806 761 267 2 × 2 = 0 + 0.796 875 009 613 522 534 4;
  • 31) 0.796 875 009 613 522 534 4 × 2 = 1 + 0.593 750 019 227 045 068 8;
  • 32) 0.593 750 019 227 045 068 8 × 2 = 1 + 0.187 500 038 454 090 137 6;
  • 33) 0.187 500 038 454 090 137 6 × 2 = 0 + 0.375 000 076 908 180 275 2;
  • 34) 0.375 000 076 908 180 275 2 × 2 = 0 + 0.750 000 153 816 360 550 4;
  • 35) 0.750 000 153 816 360 550 4 × 2 = 1 + 0.500 000 307 632 721 100 8;
  • 36) 0.500 000 307 632 721 100 8 × 2 = 1 + 0.000 000 615 265 442 201 6;
  • 37) 0.000 000 615 265 442 201 6 × 2 = 0 + 0.000 001 230 530 884 403 2;
  • 38) 0.000 001 230 530 884 403 2 × 2 = 0 + 0.000 002 461 061 768 806 4;
  • 39) 0.000 002 461 061 768 806 4 × 2 = 0 + 0.000 004 922 123 537 612 8;
  • 40) 0.000 004 922 123 537 612 8 × 2 = 0 + 0.000 009 844 247 075 225 6;
  • 41) 0.000 009 844 247 075 225 6 × 2 = 0 + 0.000 019 688 494 150 451 2;
  • 42) 0.000 019 688 494 150 451 2 × 2 = 0 + 0.000 039 376 988 300 902 4;
  • 43) 0.000 039 376 988 300 902 4 × 2 = 0 + 0.000 078 753 976 601 804 8;
  • 44) 0.000 078 753 976 601 804 8 × 2 = 0 + 0.000 157 507 953 203 609 6;
  • 45) 0.000 157 507 953 203 609 6 × 2 = 0 + 0.000 315 015 906 407 219 2;
  • 46) 0.000 315 015 906 407 219 2 × 2 = 0 + 0.000 630 031 812 814 438 4;
  • 47) 0.000 630 031 812 814 438 4 × 2 = 0 + 0.001 260 063 625 628 876 8;
  • 48) 0.001 260 063 625 628 876 8 × 2 = 0 + 0.002 520 127 251 257 753 6;
  • 49) 0.002 520 127 251 257 753 6 × 2 = 0 + 0.005 040 254 502 515 507 2;
  • 50) 0.005 040 254 502 515 507 2 × 2 = 0 + 0.010 080 509 005 031 014 4;
  • 51) 0.010 080 509 005 031 014 4 × 2 = 0 + 0.020 161 018 010 062 028 8;
  • 52) 0.020 161 018 010 062 028 8 × 2 = 0 + 0.040 322 036 020 124 057 6;
  • 53) 0.040 322 036 020 124 057 6 × 2 = 0 + 0.080 644 072 040 248 115 2;
  • 54) 0.080 644 072 040 248 115 2 × 2 = 0 + 0.161 288 144 080 496 230 4;

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:


0.000 000 000 742 147 685 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 685 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 685 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 000


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 0000 =

100 1100 0000 0000 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 0000

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


Decimal number -0.000 000 000 742 147 685 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 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