-0.000 000 000 742 147 683 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 683 4(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
-0.000 000 000 742 147 683 4(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:

|-0.000 000 000 742 147 683 4| = 0.000 000 000 742 147 683 4


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: 0.000 000 000 742 147 683 4.

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) 0.000 000 000 742 147 683 4 × 2 = 0 + 0.000 000 001 484 295 366 8;
  • 2) 0.000 000 001 484 295 366 8 × 2 = 0 + 0.000 000 002 968 590 733 6;
  • 3) 0.000 000 002 968 590 733 6 × 2 = 0 + 0.000 000 005 937 181 467 2;
  • 4) 0.000 000 005 937 181 467 2 × 2 = 0 + 0.000 000 011 874 362 934 4;
  • 5) 0.000 000 011 874 362 934 4 × 2 = 0 + 0.000 000 023 748 725 868 8;
  • 6) 0.000 000 023 748 725 868 8 × 2 = 0 + 0.000 000 047 497 451 737 6;
  • 7) 0.000 000 047 497 451 737 6 × 2 = 0 + 0.000 000 094 994 903 475 2;
  • 8) 0.000 000 094 994 903 475 2 × 2 = 0 + 0.000 000 189 989 806 950 4;
  • 9) 0.000 000 189 989 806 950 4 × 2 = 0 + 0.000 000 379 979 613 900 8;
  • 10) 0.000 000 379 979 613 900 8 × 2 = 0 + 0.000 000 759 959 227 801 6;
  • 11) 0.000 000 759 959 227 801 6 × 2 = 0 + 0.000 001 519 918 455 603 2;
  • 12) 0.000 001 519 918 455 603 2 × 2 = 0 + 0.000 003 039 836 911 206 4;
  • 13) 0.000 003 039 836 911 206 4 × 2 = 0 + 0.000 006 079 673 822 412 8;
  • 14) 0.000 006 079 673 822 412 8 × 2 = 0 + 0.000 012 159 347 644 825 6;
  • 15) 0.000 012 159 347 644 825 6 × 2 = 0 + 0.000 024 318 695 289 651 2;
  • 16) 0.000 024 318 695 289 651 2 × 2 = 0 + 0.000 048 637 390 579 302 4;
  • 17) 0.000 048 637 390 579 302 4 × 2 = 0 + 0.000 097 274 781 158 604 8;
  • 18) 0.000 097 274 781 158 604 8 × 2 = 0 + 0.000 194 549 562 317 209 6;
  • 19) 0.000 194 549 562 317 209 6 × 2 = 0 + 0.000 389 099 124 634 419 2;
  • 20) 0.000 389 099 124 634 419 2 × 2 = 0 + 0.000 778 198 249 268 838 4;
  • 21) 0.000 778 198 249 268 838 4 × 2 = 0 + 0.001 556 396 498 537 676 8;
  • 22) 0.001 556 396 498 537 676 8 × 2 = 0 + 0.003 112 792 997 075 353 6;
  • 23) 0.003 112 792 997 075 353 6 × 2 = 0 + 0.006 225 585 994 150 707 2;
  • 24) 0.006 225 585 994 150 707 2 × 2 = 0 + 0.012 451 171 988 301 414 4;
  • 25) 0.012 451 171 988 301 414 4 × 2 = 0 + 0.024 902 343 976 602 828 8;
  • 26) 0.024 902 343 976 602 828 8 × 2 = 0 + 0.049 804 687 953 205 657 6;
  • 27) 0.049 804 687 953 205 657 6 × 2 = 0 + 0.099 609 375 906 411 315 2;
  • 28) 0.099 609 375 906 411 315 2 × 2 = 0 + 0.199 218 751 812 822 630 4;
  • 29) 0.199 218 751 812 822 630 4 × 2 = 0 + 0.398 437 503 625 645 260 8;
  • 30) 0.398 437 503 625 645 260 8 × 2 = 0 + 0.796 875 007 251 290 521 6;
  • 31) 0.796 875 007 251 290 521 6 × 2 = 1 + 0.593 750 014 502 581 043 2;
  • 32) 0.593 750 014 502 581 043 2 × 2 = 1 + 0.187 500 029 005 162 086 4;
  • 33) 0.187 500 029 005 162 086 4 × 2 = 0 + 0.375 000 058 010 324 172 8;
  • 34) 0.375 000 058 010 324 172 8 × 2 = 0 + 0.750 000 116 020 648 345 6;
  • 35) 0.750 000 116 020 648 345 6 × 2 = 1 + 0.500 000 232 041 296 691 2;
  • 36) 0.500 000 232 041 296 691 2 × 2 = 1 + 0.000 000 464 082 593 382 4;
  • 37) 0.000 000 464 082 593 382 4 × 2 = 0 + 0.000 000 928 165 186 764 8;
  • 38) 0.000 000 928 165 186 764 8 × 2 = 0 + 0.000 001 856 330 373 529 6;
  • 39) 0.000 001 856 330 373 529 6 × 2 = 0 + 0.000 003 712 660 747 059 2;
  • 40) 0.000 003 712 660 747 059 2 × 2 = 0 + 0.000 007 425 321 494 118 4;
  • 41) 0.000 007 425 321 494 118 4 × 2 = 0 + 0.000 014 850 642 988 236 8;
  • 42) 0.000 014 850 642 988 236 8 × 2 = 0 + 0.000 029 701 285 976 473 6;
  • 43) 0.000 029 701 285 976 473 6 × 2 = 0 + 0.000 059 402 571 952 947 2;
  • 44) 0.000 059 402 571 952 947 2 × 2 = 0 + 0.000 118 805 143 905 894 4;
  • 45) 0.000 118 805 143 905 894 4 × 2 = 0 + 0.000 237 610 287 811 788 8;
  • 46) 0.000 237 610 287 811 788 8 × 2 = 0 + 0.000 475 220 575 623 577 6;
  • 47) 0.000 475 220 575 623 577 6 × 2 = 0 + 0.000 950 441 151 247 155 2;
  • 48) 0.000 950 441 151 247 155 2 × 2 = 0 + 0.001 900 882 302 494 310 4;
  • 49) 0.001 900 882 302 494 310 4 × 2 = 0 + 0.003 801 764 604 988 620 8;
  • 50) 0.003 801 764 604 988 620 8 × 2 = 0 + 0.007 603 529 209 977 241 6;
  • 51) 0.007 603 529 209 977 241 6 × 2 = 0 + 0.015 207 058 419 954 483 2;
  • 52) 0.015 207 058 419 954 483 2 × 2 = 0 + 0.030 414 116 839 908 966 4;
  • 53) 0.030 414 116 839 908 966 4 × 2 = 0 + 0.060 828 233 679 817 932 8;
  • 54) 0.060 828 233 679 817 932 8 × 2 = 0 + 0.121 656 467 359 635 865 6;

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:


0.000 000 000 742 147 683 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 683 4(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:


0.000 000 000 742 147 683 4(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 -0.000 000 000 742 147 683 4 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