-0.000 000 000 742 147 676 600 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 600 9(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
-0.000 000 000 742 147 676 600 9(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:

|-0.000 000 000 742 147 676 600 9| = 0.000 000 000 742 147 676 600 9


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: 0.000 000 000 742 147 676 600 9.

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) 0.000 000 000 742 147 676 600 9 × 2 = 0 + 0.000 000 001 484 295 353 201 8;
  • 2) 0.000 000 001 484 295 353 201 8 × 2 = 0 + 0.000 000 002 968 590 706 403 6;
  • 3) 0.000 000 002 968 590 706 403 6 × 2 = 0 + 0.000 000 005 937 181 412 807 2;
  • 4) 0.000 000 005 937 181 412 807 2 × 2 = 0 + 0.000 000 011 874 362 825 614 4;
  • 5) 0.000 000 011 874 362 825 614 4 × 2 = 0 + 0.000 000 023 748 725 651 228 8;
  • 6) 0.000 000 023 748 725 651 228 8 × 2 = 0 + 0.000 000 047 497 451 302 457 6;
  • 7) 0.000 000 047 497 451 302 457 6 × 2 = 0 + 0.000 000 094 994 902 604 915 2;
  • 8) 0.000 000 094 994 902 604 915 2 × 2 = 0 + 0.000 000 189 989 805 209 830 4;
  • 9) 0.000 000 189 989 805 209 830 4 × 2 = 0 + 0.000 000 379 979 610 419 660 8;
  • 10) 0.000 000 379 979 610 419 660 8 × 2 = 0 + 0.000 000 759 959 220 839 321 6;
  • 11) 0.000 000 759 959 220 839 321 6 × 2 = 0 + 0.000 001 519 918 441 678 643 2;
  • 12) 0.000 001 519 918 441 678 643 2 × 2 = 0 + 0.000 003 039 836 883 357 286 4;
  • 13) 0.000 003 039 836 883 357 286 4 × 2 = 0 + 0.000 006 079 673 766 714 572 8;
  • 14) 0.000 006 079 673 766 714 572 8 × 2 = 0 + 0.000 012 159 347 533 429 145 6;
  • 15) 0.000 012 159 347 533 429 145 6 × 2 = 0 + 0.000 024 318 695 066 858 291 2;
  • 16) 0.000 024 318 695 066 858 291 2 × 2 = 0 + 0.000 048 637 390 133 716 582 4;
  • 17) 0.000 048 637 390 133 716 582 4 × 2 = 0 + 0.000 097 274 780 267 433 164 8;
  • 18) 0.000 097 274 780 267 433 164 8 × 2 = 0 + 0.000 194 549 560 534 866 329 6;
  • 19) 0.000 194 549 560 534 866 329 6 × 2 = 0 + 0.000 389 099 121 069 732 659 2;
  • 20) 0.000 389 099 121 069 732 659 2 × 2 = 0 + 0.000 778 198 242 139 465 318 4;
  • 21) 0.000 778 198 242 139 465 318 4 × 2 = 0 + 0.001 556 396 484 278 930 636 8;
  • 22) 0.001 556 396 484 278 930 636 8 × 2 = 0 + 0.003 112 792 968 557 861 273 6;
  • 23) 0.003 112 792 968 557 861 273 6 × 2 = 0 + 0.006 225 585 937 115 722 547 2;
  • 24) 0.006 225 585 937 115 722 547 2 × 2 = 0 + 0.012 451 171 874 231 445 094 4;
  • 25) 0.012 451 171 874 231 445 094 4 × 2 = 0 + 0.024 902 343 748 462 890 188 8;
  • 26) 0.024 902 343 748 462 890 188 8 × 2 = 0 + 0.049 804 687 496 925 780 377 6;
  • 27) 0.049 804 687 496 925 780 377 6 × 2 = 0 + 0.099 609 374 993 851 560 755 2;
  • 28) 0.099 609 374 993 851 560 755 2 × 2 = 0 + 0.199 218 749 987 703 121 510 4;
  • 29) 0.199 218 749 987 703 121 510 4 × 2 = 0 + 0.398 437 499 975 406 243 020 8;
  • 30) 0.398 437 499 975 406 243 020 8 × 2 = 0 + 0.796 874 999 950 812 486 041 6;
  • 31) 0.796 874 999 950 812 486 041 6 × 2 = 1 + 0.593 749 999 901 624 972 083 2;
  • 32) 0.593 749 999 901 624 972 083 2 × 2 = 1 + 0.187 499 999 803 249 944 166 4;
  • 33) 0.187 499 999 803 249 944 166 4 × 2 = 0 + 0.374 999 999 606 499 888 332 8;
  • 34) 0.374 999 999 606 499 888 332 8 × 2 = 0 + 0.749 999 999 212 999 776 665 6;
  • 35) 0.749 999 999 212 999 776 665 6 × 2 = 1 + 0.499 999 998 425 999 553 331 2;
  • 36) 0.499 999 998 425 999 553 331 2 × 2 = 0 + 0.999 999 996 851 999 106 662 4;
  • 37) 0.999 999 996 851 999 106 662 4 × 2 = 1 + 0.999 999 993 703 998 213 324 8;
  • 38) 0.999 999 993 703 998 213 324 8 × 2 = 1 + 0.999 999 987 407 996 426 649 6;
  • 39) 0.999 999 987 407 996 426 649 6 × 2 = 1 + 0.999 999 974 815 992 853 299 2;
  • 40) 0.999 999 974 815 992 853 299 2 × 2 = 1 + 0.999 999 949 631 985 706 598 4;
  • 41) 0.999 999 949 631 985 706 598 4 × 2 = 1 + 0.999 999 899 263 971 413 196 8;
  • 42) 0.999 999 899 263 971 413 196 8 × 2 = 1 + 0.999 999 798 527 942 826 393 6;
  • 43) 0.999 999 798 527 942 826 393 6 × 2 = 1 + 0.999 999 597 055 885 652 787 2;
  • 44) 0.999 999 597 055 885 652 787 2 × 2 = 1 + 0.999 999 194 111 771 305 574 4;
  • 45) 0.999 999 194 111 771 305 574 4 × 2 = 1 + 0.999 998 388 223 542 611 148 8;
  • 46) 0.999 998 388 223 542 611 148 8 × 2 = 1 + 0.999 996 776 447 085 222 297 6;
  • 47) 0.999 996 776 447 085 222 297 6 × 2 = 1 + 0.999 993 552 894 170 444 595 2;
  • 48) 0.999 993 552 894 170 444 595 2 × 2 = 1 + 0.999 987 105 788 340 889 190 4;
  • 49) 0.999 987 105 788 340 889 190 4 × 2 = 1 + 0.999 974 211 576 681 778 380 8;
  • 50) 0.999 974 211 576 681 778 380 8 × 2 = 1 + 0.999 948 423 153 363 556 761 6;
  • 51) 0.999 948 423 153 363 556 761 6 × 2 = 1 + 0.999 896 846 306 727 113 523 2;
  • 52) 0.999 896 846 306 727 113 523 2 × 2 = 1 + 0.999 793 692 613 454 227 046 4;
  • 53) 0.999 793 692 613 454 227 046 4 × 2 = 1 + 0.999 587 385 226 908 454 092 8;
  • 54) 0.999 587 385 226 908 454 092 8 × 2 = 1 + 0.999 174 770 453 816 908 185 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:


0.000 000 000 742 147 676 600 9(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 600 9(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:


0.000 000 000 742 147 676 600 9(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 -0.000 000 000 742 147 676 600 9 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