-0.000 000 000 742 147 676 648 56 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648 56(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
-0.000 000 000 742 147 676 648 56(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:

|-0.000 000 000 742 147 676 648 56| = 0.000 000 000 742 147 676 648 56


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: 0.000 000 000 742 147 676 648 56.

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) 0.000 000 000 742 147 676 648 56 × 2 = 0 + 0.000 000 001 484 295 353 297 12;
  • 2) 0.000 000 001 484 295 353 297 12 × 2 = 0 + 0.000 000 002 968 590 706 594 24;
  • 3) 0.000 000 002 968 590 706 594 24 × 2 = 0 + 0.000 000 005 937 181 413 188 48;
  • 4) 0.000 000 005 937 181 413 188 48 × 2 = 0 + 0.000 000 011 874 362 826 376 96;
  • 5) 0.000 000 011 874 362 826 376 96 × 2 = 0 + 0.000 000 023 748 725 652 753 92;
  • 6) 0.000 000 023 748 725 652 753 92 × 2 = 0 + 0.000 000 047 497 451 305 507 84;
  • 7) 0.000 000 047 497 451 305 507 84 × 2 = 0 + 0.000 000 094 994 902 611 015 68;
  • 8) 0.000 000 094 994 902 611 015 68 × 2 = 0 + 0.000 000 189 989 805 222 031 36;
  • 9) 0.000 000 189 989 805 222 031 36 × 2 = 0 + 0.000 000 379 979 610 444 062 72;
  • 10) 0.000 000 379 979 610 444 062 72 × 2 = 0 + 0.000 000 759 959 220 888 125 44;
  • 11) 0.000 000 759 959 220 888 125 44 × 2 = 0 + 0.000 001 519 918 441 776 250 88;
  • 12) 0.000 001 519 918 441 776 250 88 × 2 = 0 + 0.000 003 039 836 883 552 501 76;
  • 13) 0.000 003 039 836 883 552 501 76 × 2 = 0 + 0.000 006 079 673 767 105 003 52;
  • 14) 0.000 006 079 673 767 105 003 52 × 2 = 0 + 0.000 012 159 347 534 210 007 04;
  • 15) 0.000 012 159 347 534 210 007 04 × 2 = 0 + 0.000 024 318 695 068 420 014 08;
  • 16) 0.000 024 318 695 068 420 014 08 × 2 = 0 + 0.000 048 637 390 136 840 028 16;
  • 17) 0.000 048 637 390 136 840 028 16 × 2 = 0 + 0.000 097 274 780 273 680 056 32;
  • 18) 0.000 097 274 780 273 680 056 32 × 2 = 0 + 0.000 194 549 560 547 360 112 64;
  • 19) 0.000 194 549 560 547 360 112 64 × 2 = 0 + 0.000 389 099 121 094 720 225 28;
  • 20) 0.000 389 099 121 094 720 225 28 × 2 = 0 + 0.000 778 198 242 189 440 450 56;
  • 21) 0.000 778 198 242 189 440 450 56 × 2 = 0 + 0.001 556 396 484 378 880 901 12;
  • 22) 0.001 556 396 484 378 880 901 12 × 2 = 0 + 0.003 112 792 968 757 761 802 24;
  • 23) 0.003 112 792 968 757 761 802 24 × 2 = 0 + 0.006 225 585 937 515 523 604 48;
  • 24) 0.006 225 585 937 515 523 604 48 × 2 = 0 + 0.012 451 171 875 031 047 208 96;
  • 25) 0.012 451 171 875 031 047 208 96 × 2 = 0 + 0.024 902 343 750 062 094 417 92;
  • 26) 0.024 902 343 750 062 094 417 92 × 2 = 0 + 0.049 804 687 500 124 188 835 84;
  • 27) 0.049 804 687 500 124 188 835 84 × 2 = 0 + 0.099 609 375 000 248 377 671 68;
  • 28) 0.099 609 375 000 248 377 671 68 × 2 = 0 + 0.199 218 750 000 496 755 343 36;
  • 29) 0.199 218 750 000 496 755 343 36 × 2 = 0 + 0.398 437 500 000 993 510 686 72;
  • 30) 0.398 437 500 000 993 510 686 72 × 2 = 0 + 0.796 875 000 001 987 021 373 44;
  • 31) 0.796 875 000 001 987 021 373 44 × 2 = 1 + 0.593 750 000 003 974 042 746 88;
  • 32) 0.593 750 000 003 974 042 746 88 × 2 = 1 + 0.187 500 000 007 948 085 493 76;
  • 33) 0.187 500 000 007 948 085 493 76 × 2 = 0 + 0.375 000 000 015 896 170 987 52;
  • 34) 0.375 000 000 015 896 170 987 52 × 2 = 0 + 0.750 000 000 031 792 341 975 04;
  • 35) 0.750 000 000 031 792 341 975 04 × 2 = 1 + 0.500 000 000 063 584 683 950 08;
  • 36) 0.500 000 000 063 584 683 950 08 × 2 = 1 + 0.000 000 000 127 169 367 900 16;
  • 37) 0.000 000 000 127 169 367 900 16 × 2 = 0 + 0.000 000 000 254 338 735 800 32;
  • 38) 0.000 000 000 254 338 735 800 32 × 2 = 0 + 0.000 000 000 508 677 471 600 64;
  • 39) 0.000 000 000 508 677 471 600 64 × 2 = 0 + 0.000 000 001 017 354 943 201 28;
  • 40) 0.000 000 001 017 354 943 201 28 × 2 = 0 + 0.000 000 002 034 709 886 402 56;
  • 41) 0.000 000 002 034 709 886 402 56 × 2 = 0 + 0.000 000 004 069 419 772 805 12;
  • 42) 0.000 000 004 069 419 772 805 12 × 2 = 0 + 0.000 000 008 138 839 545 610 24;
  • 43) 0.000 000 008 138 839 545 610 24 × 2 = 0 + 0.000 000 016 277 679 091 220 48;
  • 44) 0.000 000 016 277 679 091 220 48 × 2 = 0 + 0.000 000 032 555 358 182 440 96;
  • 45) 0.000 000 032 555 358 182 440 96 × 2 = 0 + 0.000 000 065 110 716 364 881 92;
  • 46) 0.000 000 065 110 716 364 881 92 × 2 = 0 + 0.000 000 130 221 432 729 763 84;
  • 47) 0.000 000 130 221 432 729 763 84 × 2 = 0 + 0.000 000 260 442 865 459 527 68;
  • 48) 0.000 000 260 442 865 459 527 68 × 2 = 0 + 0.000 000 520 885 730 919 055 36;
  • 49) 0.000 000 520 885 730 919 055 36 × 2 = 0 + 0.000 001 041 771 461 838 110 72;
  • 50) 0.000 001 041 771 461 838 110 72 × 2 = 0 + 0.000 002 083 542 923 676 221 44;
  • 51) 0.000 002 083 542 923 676 221 44 × 2 = 0 + 0.000 004 167 085 847 352 442 88;
  • 52) 0.000 004 167 085 847 352 442 88 × 2 = 0 + 0.000 008 334 171 694 704 885 76;
  • 53) 0.000 008 334 171 694 704 885 76 × 2 = 0 + 0.000 016 668 343 389 409 771 52;
  • 54) 0.000 016 668 343 389 409 771 52 × 2 = 0 + 0.000 033 336 686 778 819 543 04;

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:


0.000 000 000 742 147 676 648 56(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 648 56(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:


0.000 000 000 742 147 676 648 56(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 -0.000 000 000 742 147 676 648 56 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