-0.000 000 000 742 147 676 645 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 645 93(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
-0.000 000 000 742 147 676 645 93(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:

|-0.000 000 000 742 147 676 645 93| = 0.000 000 000 742 147 676 645 93


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: 0.000 000 000 742 147 676 645 93.

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) 0.000 000 000 742 147 676 645 93 × 2 = 0 + 0.000 000 001 484 295 353 291 86;
  • 2) 0.000 000 001 484 295 353 291 86 × 2 = 0 + 0.000 000 002 968 590 706 583 72;
  • 3) 0.000 000 002 968 590 706 583 72 × 2 = 0 + 0.000 000 005 937 181 413 167 44;
  • 4) 0.000 000 005 937 181 413 167 44 × 2 = 0 + 0.000 000 011 874 362 826 334 88;
  • 5) 0.000 000 011 874 362 826 334 88 × 2 = 0 + 0.000 000 023 748 725 652 669 76;
  • 6) 0.000 000 023 748 725 652 669 76 × 2 = 0 + 0.000 000 047 497 451 305 339 52;
  • 7) 0.000 000 047 497 451 305 339 52 × 2 = 0 + 0.000 000 094 994 902 610 679 04;
  • 8) 0.000 000 094 994 902 610 679 04 × 2 = 0 + 0.000 000 189 989 805 221 358 08;
  • 9) 0.000 000 189 989 805 221 358 08 × 2 = 0 + 0.000 000 379 979 610 442 716 16;
  • 10) 0.000 000 379 979 610 442 716 16 × 2 = 0 + 0.000 000 759 959 220 885 432 32;
  • 11) 0.000 000 759 959 220 885 432 32 × 2 = 0 + 0.000 001 519 918 441 770 864 64;
  • 12) 0.000 001 519 918 441 770 864 64 × 2 = 0 + 0.000 003 039 836 883 541 729 28;
  • 13) 0.000 003 039 836 883 541 729 28 × 2 = 0 + 0.000 006 079 673 767 083 458 56;
  • 14) 0.000 006 079 673 767 083 458 56 × 2 = 0 + 0.000 012 159 347 534 166 917 12;
  • 15) 0.000 012 159 347 534 166 917 12 × 2 = 0 + 0.000 024 318 695 068 333 834 24;
  • 16) 0.000 024 318 695 068 333 834 24 × 2 = 0 + 0.000 048 637 390 136 667 668 48;
  • 17) 0.000 048 637 390 136 667 668 48 × 2 = 0 + 0.000 097 274 780 273 335 336 96;
  • 18) 0.000 097 274 780 273 335 336 96 × 2 = 0 + 0.000 194 549 560 546 670 673 92;
  • 19) 0.000 194 549 560 546 670 673 92 × 2 = 0 + 0.000 389 099 121 093 341 347 84;
  • 20) 0.000 389 099 121 093 341 347 84 × 2 = 0 + 0.000 778 198 242 186 682 695 68;
  • 21) 0.000 778 198 242 186 682 695 68 × 2 = 0 + 0.001 556 396 484 373 365 391 36;
  • 22) 0.001 556 396 484 373 365 391 36 × 2 = 0 + 0.003 112 792 968 746 730 782 72;
  • 23) 0.003 112 792 968 746 730 782 72 × 2 = 0 + 0.006 225 585 937 493 461 565 44;
  • 24) 0.006 225 585 937 493 461 565 44 × 2 = 0 + 0.012 451 171 874 986 923 130 88;
  • 25) 0.012 451 171 874 986 923 130 88 × 2 = 0 + 0.024 902 343 749 973 846 261 76;
  • 26) 0.024 902 343 749 973 846 261 76 × 2 = 0 + 0.049 804 687 499 947 692 523 52;
  • 27) 0.049 804 687 499 947 692 523 52 × 2 = 0 + 0.099 609 374 999 895 385 047 04;
  • 28) 0.099 609 374 999 895 385 047 04 × 2 = 0 + 0.199 218 749 999 790 770 094 08;
  • 29) 0.199 218 749 999 790 770 094 08 × 2 = 0 + 0.398 437 499 999 581 540 188 16;
  • 30) 0.398 437 499 999 581 540 188 16 × 2 = 0 + 0.796 874 999 999 163 080 376 32;
  • 31) 0.796 874 999 999 163 080 376 32 × 2 = 1 + 0.593 749 999 998 326 160 752 64;
  • 32) 0.593 749 999 998 326 160 752 64 × 2 = 1 + 0.187 499 999 996 652 321 505 28;
  • 33) 0.187 499 999 996 652 321 505 28 × 2 = 0 + 0.374 999 999 993 304 643 010 56;
  • 34) 0.374 999 999 993 304 643 010 56 × 2 = 0 + 0.749 999 999 986 609 286 021 12;
  • 35) 0.749 999 999 986 609 286 021 12 × 2 = 1 + 0.499 999 999 973 218 572 042 24;
  • 36) 0.499 999 999 973 218 572 042 24 × 2 = 0 + 0.999 999 999 946 437 144 084 48;
  • 37) 0.999 999 999 946 437 144 084 48 × 2 = 1 + 0.999 999 999 892 874 288 168 96;
  • 38) 0.999 999 999 892 874 288 168 96 × 2 = 1 + 0.999 999 999 785 748 576 337 92;
  • 39) 0.999 999 999 785 748 576 337 92 × 2 = 1 + 0.999 999 999 571 497 152 675 84;
  • 40) 0.999 999 999 571 497 152 675 84 × 2 = 1 + 0.999 999 999 142 994 305 351 68;
  • 41) 0.999 999 999 142 994 305 351 68 × 2 = 1 + 0.999 999 998 285 988 610 703 36;
  • 42) 0.999 999 998 285 988 610 703 36 × 2 = 1 + 0.999 999 996 571 977 221 406 72;
  • 43) 0.999 999 996 571 977 221 406 72 × 2 = 1 + 0.999 999 993 143 954 442 813 44;
  • 44) 0.999 999 993 143 954 442 813 44 × 2 = 1 + 0.999 999 986 287 908 885 626 88;
  • 45) 0.999 999 986 287 908 885 626 88 × 2 = 1 + 0.999 999 972 575 817 771 253 76;
  • 46) 0.999 999 972 575 817 771 253 76 × 2 = 1 + 0.999 999 945 151 635 542 507 52;
  • 47) 0.999 999 945 151 635 542 507 52 × 2 = 1 + 0.999 999 890 303 271 085 015 04;
  • 48) 0.999 999 890 303 271 085 015 04 × 2 = 1 + 0.999 999 780 606 542 170 030 08;
  • 49) 0.999 999 780 606 542 170 030 08 × 2 = 1 + 0.999 999 561 213 084 340 060 16;
  • 50) 0.999 999 561 213 084 340 060 16 × 2 = 1 + 0.999 999 122 426 168 680 120 32;
  • 51) 0.999 999 122 426 168 680 120 32 × 2 = 1 + 0.999 998 244 852 337 360 240 64;
  • 52) 0.999 998 244 852 337 360 240 64 × 2 = 1 + 0.999 996 489 704 674 720 481 28;
  • 53) 0.999 996 489 704 674 720 481 28 × 2 = 1 + 0.999 992 979 409 349 440 962 56;
  • 54) 0.999 992 979 409 349 440 962 56 × 2 = 1 + 0.999 985 958 818 698 881 925 12;

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:


0.000 000 000 742 147 676 645 93(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 645 93(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:


0.000 000 000 742 147 676 645 93(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 -0.000 000 000 742 147 676 645 93 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 645 52 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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