-0.000 000 000 742 147 676 648 79 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648 79(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 79(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 79| = 0.000 000 000 742 147 676 648 79


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 79.

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 79 × 2 = 0 + 0.000 000 001 484 295 353 297 58;
  • 2) 0.000 000 001 484 295 353 297 58 × 2 = 0 + 0.000 000 002 968 590 706 595 16;
  • 3) 0.000 000 002 968 590 706 595 16 × 2 = 0 + 0.000 000 005 937 181 413 190 32;
  • 4) 0.000 000 005 937 181 413 190 32 × 2 = 0 + 0.000 000 011 874 362 826 380 64;
  • 5) 0.000 000 011 874 362 826 380 64 × 2 = 0 + 0.000 000 023 748 725 652 761 28;
  • 6) 0.000 000 023 748 725 652 761 28 × 2 = 0 + 0.000 000 047 497 451 305 522 56;
  • 7) 0.000 000 047 497 451 305 522 56 × 2 = 0 + 0.000 000 094 994 902 611 045 12;
  • 8) 0.000 000 094 994 902 611 045 12 × 2 = 0 + 0.000 000 189 989 805 222 090 24;
  • 9) 0.000 000 189 989 805 222 090 24 × 2 = 0 + 0.000 000 379 979 610 444 180 48;
  • 10) 0.000 000 379 979 610 444 180 48 × 2 = 0 + 0.000 000 759 959 220 888 360 96;
  • 11) 0.000 000 759 959 220 888 360 96 × 2 = 0 + 0.000 001 519 918 441 776 721 92;
  • 12) 0.000 001 519 918 441 776 721 92 × 2 = 0 + 0.000 003 039 836 883 553 443 84;
  • 13) 0.000 003 039 836 883 553 443 84 × 2 = 0 + 0.000 006 079 673 767 106 887 68;
  • 14) 0.000 006 079 673 767 106 887 68 × 2 = 0 + 0.000 012 159 347 534 213 775 36;
  • 15) 0.000 012 159 347 534 213 775 36 × 2 = 0 + 0.000 024 318 695 068 427 550 72;
  • 16) 0.000 024 318 695 068 427 550 72 × 2 = 0 + 0.000 048 637 390 136 855 101 44;
  • 17) 0.000 048 637 390 136 855 101 44 × 2 = 0 + 0.000 097 274 780 273 710 202 88;
  • 18) 0.000 097 274 780 273 710 202 88 × 2 = 0 + 0.000 194 549 560 547 420 405 76;
  • 19) 0.000 194 549 560 547 420 405 76 × 2 = 0 + 0.000 389 099 121 094 840 811 52;
  • 20) 0.000 389 099 121 094 840 811 52 × 2 = 0 + 0.000 778 198 242 189 681 623 04;
  • 21) 0.000 778 198 242 189 681 623 04 × 2 = 0 + 0.001 556 396 484 379 363 246 08;
  • 22) 0.001 556 396 484 379 363 246 08 × 2 = 0 + 0.003 112 792 968 758 726 492 16;
  • 23) 0.003 112 792 968 758 726 492 16 × 2 = 0 + 0.006 225 585 937 517 452 984 32;
  • 24) 0.006 225 585 937 517 452 984 32 × 2 = 0 + 0.012 451 171 875 034 905 968 64;
  • 25) 0.012 451 171 875 034 905 968 64 × 2 = 0 + 0.024 902 343 750 069 811 937 28;
  • 26) 0.024 902 343 750 069 811 937 28 × 2 = 0 + 0.049 804 687 500 139 623 874 56;
  • 27) 0.049 804 687 500 139 623 874 56 × 2 = 0 + 0.099 609 375 000 279 247 749 12;
  • 28) 0.099 609 375 000 279 247 749 12 × 2 = 0 + 0.199 218 750 000 558 495 498 24;
  • 29) 0.199 218 750 000 558 495 498 24 × 2 = 0 + 0.398 437 500 001 116 990 996 48;
  • 30) 0.398 437 500 001 116 990 996 48 × 2 = 0 + 0.796 875 000 002 233 981 992 96;
  • 31) 0.796 875 000 002 233 981 992 96 × 2 = 1 + 0.593 750 000 004 467 963 985 92;
  • 32) 0.593 750 000 004 467 963 985 92 × 2 = 1 + 0.187 500 000 008 935 927 971 84;
  • 33) 0.187 500 000 008 935 927 971 84 × 2 = 0 + 0.375 000 000 017 871 855 943 68;
  • 34) 0.375 000 000 017 871 855 943 68 × 2 = 0 + 0.750 000 000 035 743 711 887 36;
  • 35) 0.750 000 000 035 743 711 887 36 × 2 = 1 + 0.500 000 000 071 487 423 774 72;
  • 36) 0.500 000 000 071 487 423 774 72 × 2 = 1 + 0.000 000 000 142 974 847 549 44;
  • 37) 0.000 000 000 142 974 847 549 44 × 2 = 0 + 0.000 000 000 285 949 695 098 88;
  • 38) 0.000 000 000 285 949 695 098 88 × 2 = 0 + 0.000 000 000 571 899 390 197 76;
  • 39) 0.000 000 000 571 899 390 197 76 × 2 = 0 + 0.000 000 001 143 798 780 395 52;
  • 40) 0.000 000 001 143 798 780 395 52 × 2 = 0 + 0.000 000 002 287 597 560 791 04;
  • 41) 0.000 000 002 287 597 560 791 04 × 2 = 0 + 0.000 000 004 575 195 121 582 08;
  • 42) 0.000 000 004 575 195 121 582 08 × 2 = 0 + 0.000 000 009 150 390 243 164 16;
  • 43) 0.000 000 009 150 390 243 164 16 × 2 = 0 + 0.000 000 018 300 780 486 328 32;
  • 44) 0.000 000 018 300 780 486 328 32 × 2 = 0 + 0.000 000 036 601 560 972 656 64;
  • 45) 0.000 000 036 601 560 972 656 64 × 2 = 0 + 0.000 000 073 203 121 945 313 28;
  • 46) 0.000 000 073 203 121 945 313 28 × 2 = 0 + 0.000 000 146 406 243 890 626 56;
  • 47) 0.000 000 146 406 243 890 626 56 × 2 = 0 + 0.000 000 292 812 487 781 253 12;
  • 48) 0.000 000 292 812 487 781 253 12 × 2 = 0 + 0.000 000 585 624 975 562 506 24;
  • 49) 0.000 000 585 624 975 562 506 24 × 2 = 0 + 0.000 001 171 249 951 125 012 48;
  • 50) 0.000 001 171 249 951 125 012 48 × 2 = 0 + 0.000 002 342 499 902 250 024 96;
  • 51) 0.000 002 342 499 902 250 024 96 × 2 = 0 + 0.000 004 684 999 804 500 049 92;
  • 52) 0.000 004 684 999 804 500 049 92 × 2 = 0 + 0.000 009 369 999 609 000 099 84;
  • 53) 0.000 009 369 999 609 000 099 84 × 2 = 0 + 0.000 018 739 999 218 000 199 68;
  • 54) 0.000 018 739 999 218 000 199 68 × 2 = 0 + 0.000 037 479 998 436 000 399 36;

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 79(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 79(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 79(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 79 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