-0.000 000 000 742 147 676 646 52 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 52(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
-0.000 000 000 742 147 676 646 52(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:

|-0.000 000 000 742 147 676 646 52| = 0.000 000 000 742 147 676 646 52


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: 0.000 000 000 742 147 676 646 52.

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) 0.000 000 000 742 147 676 646 52 × 2 = 0 + 0.000 000 001 484 295 353 293 04;
  • 2) 0.000 000 001 484 295 353 293 04 × 2 = 0 + 0.000 000 002 968 590 706 586 08;
  • 3) 0.000 000 002 968 590 706 586 08 × 2 = 0 + 0.000 000 005 937 181 413 172 16;
  • 4) 0.000 000 005 937 181 413 172 16 × 2 = 0 + 0.000 000 011 874 362 826 344 32;
  • 5) 0.000 000 011 874 362 826 344 32 × 2 = 0 + 0.000 000 023 748 725 652 688 64;
  • 6) 0.000 000 023 748 725 652 688 64 × 2 = 0 + 0.000 000 047 497 451 305 377 28;
  • 7) 0.000 000 047 497 451 305 377 28 × 2 = 0 + 0.000 000 094 994 902 610 754 56;
  • 8) 0.000 000 094 994 902 610 754 56 × 2 = 0 + 0.000 000 189 989 805 221 509 12;
  • 9) 0.000 000 189 989 805 221 509 12 × 2 = 0 + 0.000 000 379 979 610 443 018 24;
  • 10) 0.000 000 379 979 610 443 018 24 × 2 = 0 + 0.000 000 759 959 220 886 036 48;
  • 11) 0.000 000 759 959 220 886 036 48 × 2 = 0 + 0.000 001 519 918 441 772 072 96;
  • 12) 0.000 001 519 918 441 772 072 96 × 2 = 0 + 0.000 003 039 836 883 544 145 92;
  • 13) 0.000 003 039 836 883 544 145 92 × 2 = 0 + 0.000 006 079 673 767 088 291 84;
  • 14) 0.000 006 079 673 767 088 291 84 × 2 = 0 + 0.000 012 159 347 534 176 583 68;
  • 15) 0.000 012 159 347 534 176 583 68 × 2 = 0 + 0.000 024 318 695 068 353 167 36;
  • 16) 0.000 024 318 695 068 353 167 36 × 2 = 0 + 0.000 048 637 390 136 706 334 72;
  • 17) 0.000 048 637 390 136 706 334 72 × 2 = 0 + 0.000 097 274 780 273 412 669 44;
  • 18) 0.000 097 274 780 273 412 669 44 × 2 = 0 + 0.000 194 549 560 546 825 338 88;
  • 19) 0.000 194 549 560 546 825 338 88 × 2 = 0 + 0.000 389 099 121 093 650 677 76;
  • 20) 0.000 389 099 121 093 650 677 76 × 2 = 0 + 0.000 778 198 242 187 301 355 52;
  • 21) 0.000 778 198 242 187 301 355 52 × 2 = 0 + 0.001 556 396 484 374 602 711 04;
  • 22) 0.001 556 396 484 374 602 711 04 × 2 = 0 + 0.003 112 792 968 749 205 422 08;
  • 23) 0.003 112 792 968 749 205 422 08 × 2 = 0 + 0.006 225 585 937 498 410 844 16;
  • 24) 0.006 225 585 937 498 410 844 16 × 2 = 0 + 0.012 451 171 874 996 821 688 32;
  • 25) 0.012 451 171 874 996 821 688 32 × 2 = 0 + 0.024 902 343 749 993 643 376 64;
  • 26) 0.024 902 343 749 993 643 376 64 × 2 = 0 + 0.049 804 687 499 987 286 753 28;
  • 27) 0.049 804 687 499 987 286 753 28 × 2 = 0 + 0.099 609 374 999 974 573 506 56;
  • 28) 0.099 609 374 999 974 573 506 56 × 2 = 0 + 0.199 218 749 999 949 147 013 12;
  • 29) 0.199 218 749 999 949 147 013 12 × 2 = 0 + 0.398 437 499 999 898 294 026 24;
  • 30) 0.398 437 499 999 898 294 026 24 × 2 = 0 + 0.796 874 999 999 796 588 052 48;
  • 31) 0.796 874 999 999 796 588 052 48 × 2 = 1 + 0.593 749 999 999 593 176 104 96;
  • 32) 0.593 749 999 999 593 176 104 96 × 2 = 1 + 0.187 499 999 999 186 352 209 92;
  • 33) 0.187 499 999 999 186 352 209 92 × 2 = 0 + 0.374 999 999 998 372 704 419 84;
  • 34) 0.374 999 999 998 372 704 419 84 × 2 = 0 + 0.749 999 999 996 745 408 839 68;
  • 35) 0.749 999 999 996 745 408 839 68 × 2 = 1 + 0.499 999 999 993 490 817 679 36;
  • 36) 0.499 999 999 993 490 817 679 36 × 2 = 0 + 0.999 999 999 986 981 635 358 72;
  • 37) 0.999 999 999 986 981 635 358 72 × 2 = 1 + 0.999 999 999 973 963 270 717 44;
  • 38) 0.999 999 999 973 963 270 717 44 × 2 = 1 + 0.999 999 999 947 926 541 434 88;
  • 39) 0.999 999 999 947 926 541 434 88 × 2 = 1 + 0.999 999 999 895 853 082 869 76;
  • 40) 0.999 999 999 895 853 082 869 76 × 2 = 1 + 0.999 999 999 791 706 165 739 52;
  • 41) 0.999 999 999 791 706 165 739 52 × 2 = 1 + 0.999 999 999 583 412 331 479 04;
  • 42) 0.999 999 999 583 412 331 479 04 × 2 = 1 + 0.999 999 999 166 824 662 958 08;
  • 43) 0.999 999 999 166 824 662 958 08 × 2 = 1 + 0.999 999 998 333 649 325 916 16;
  • 44) 0.999 999 998 333 649 325 916 16 × 2 = 1 + 0.999 999 996 667 298 651 832 32;
  • 45) 0.999 999 996 667 298 651 832 32 × 2 = 1 + 0.999 999 993 334 597 303 664 64;
  • 46) 0.999 999 993 334 597 303 664 64 × 2 = 1 + 0.999 999 986 669 194 607 329 28;
  • 47) 0.999 999 986 669 194 607 329 28 × 2 = 1 + 0.999 999 973 338 389 214 658 56;
  • 48) 0.999 999 973 338 389 214 658 56 × 2 = 1 + 0.999 999 946 676 778 429 317 12;
  • 49) 0.999 999 946 676 778 429 317 12 × 2 = 1 + 0.999 999 893 353 556 858 634 24;
  • 50) 0.999 999 893 353 556 858 634 24 × 2 = 1 + 0.999 999 786 707 113 717 268 48;
  • 51) 0.999 999 786 707 113 717 268 48 × 2 = 1 + 0.999 999 573 414 227 434 536 96;
  • 52) 0.999 999 573 414 227 434 536 96 × 2 = 1 + 0.999 999 146 828 454 869 073 92;
  • 53) 0.999 999 146 828 454 869 073 92 × 2 = 1 + 0.999 998 293 656 909 738 147 84;
  • 54) 0.999 998 293 656 909 738 147 84 × 2 = 1 + 0.999 996 587 313 819 476 295 68;

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:


0.000 000 000 742 147 676 646 52(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 52(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:


0.000 000 000 742 147 676 646 52(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 -0.000 000 000 742 147 676 646 52 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