-0.000 000 000 742 147 676 467 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 467(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
-0.000 000 000 742 147 676 467(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:

|-0.000 000 000 742 147 676 467| = 0.000 000 000 742 147 676 467


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: 0.000 000 000 742 147 676 467.

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) 0.000 000 000 742 147 676 467 × 2 = 0 + 0.000 000 001 484 295 352 934;
  • 2) 0.000 000 001 484 295 352 934 × 2 = 0 + 0.000 000 002 968 590 705 868;
  • 3) 0.000 000 002 968 590 705 868 × 2 = 0 + 0.000 000 005 937 181 411 736;
  • 4) 0.000 000 005 937 181 411 736 × 2 = 0 + 0.000 000 011 874 362 823 472;
  • 5) 0.000 000 011 874 362 823 472 × 2 = 0 + 0.000 000 023 748 725 646 944;
  • 6) 0.000 000 023 748 725 646 944 × 2 = 0 + 0.000 000 047 497 451 293 888;
  • 7) 0.000 000 047 497 451 293 888 × 2 = 0 + 0.000 000 094 994 902 587 776;
  • 8) 0.000 000 094 994 902 587 776 × 2 = 0 + 0.000 000 189 989 805 175 552;
  • 9) 0.000 000 189 989 805 175 552 × 2 = 0 + 0.000 000 379 979 610 351 104;
  • 10) 0.000 000 379 979 610 351 104 × 2 = 0 + 0.000 000 759 959 220 702 208;
  • 11) 0.000 000 759 959 220 702 208 × 2 = 0 + 0.000 001 519 918 441 404 416;
  • 12) 0.000 001 519 918 441 404 416 × 2 = 0 + 0.000 003 039 836 882 808 832;
  • 13) 0.000 003 039 836 882 808 832 × 2 = 0 + 0.000 006 079 673 765 617 664;
  • 14) 0.000 006 079 673 765 617 664 × 2 = 0 + 0.000 012 159 347 531 235 328;
  • 15) 0.000 012 159 347 531 235 328 × 2 = 0 + 0.000 024 318 695 062 470 656;
  • 16) 0.000 024 318 695 062 470 656 × 2 = 0 + 0.000 048 637 390 124 941 312;
  • 17) 0.000 048 637 390 124 941 312 × 2 = 0 + 0.000 097 274 780 249 882 624;
  • 18) 0.000 097 274 780 249 882 624 × 2 = 0 + 0.000 194 549 560 499 765 248;
  • 19) 0.000 194 549 560 499 765 248 × 2 = 0 + 0.000 389 099 120 999 530 496;
  • 20) 0.000 389 099 120 999 530 496 × 2 = 0 + 0.000 778 198 241 999 060 992;
  • 21) 0.000 778 198 241 999 060 992 × 2 = 0 + 0.001 556 396 483 998 121 984;
  • 22) 0.001 556 396 483 998 121 984 × 2 = 0 + 0.003 112 792 967 996 243 968;
  • 23) 0.003 112 792 967 996 243 968 × 2 = 0 + 0.006 225 585 935 992 487 936;
  • 24) 0.006 225 585 935 992 487 936 × 2 = 0 + 0.012 451 171 871 984 975 872;
  • 25) 0.012 451 171 871 984 975 872 × 2 = 0 + 0.024 902 343 743 969 951 744;
  • 26) 0.024 902 343 743 969 951 744 × 2 = 0 + 0.049 804 687 487 939 903 488;
  • 27) 0.049 804 687 487 939 903 488 × 2 = 0 + 0.099 609 374 975 879 806 976;
  • 28) 0.099 609 374 975 879 806 976 × 2 = 0 + 0.199 218 749 951 759 613 952;
  • 29) 0.199 218 749 951 759 613 952 × 2 = 0 + 0.398 437 499 903 519 227 904;
  • 30) 0.398 437 499 903 519 227 904 × 2 = 0 + 0.796 874 999 807 038 455 808;
  • 31) 0.796 874 999 807 038 455 808 × 2 = 1 + 0.593 749 999 614 076 911 616;
  • 32) 0.593 749 999 614 076 911 616 × 2 = 1 + 0.187 499 999 228 153 823 232;
  • 33) 0.187 499 999 228 153 823 232 × 2 = 0 + 0.374 999 998 456 307 646 464;
  • 34) 0.374 999 998 456 307 646 464 × 2 = 0 + 0.749 999 996 912 615 292 928;
  • 35) 0.749 999 996 912 615 292 928 × 2 = 1 + 0.499 999 993 825 230 585 856;
  • 36) 0.499 999 993 825 230 585 856 × 2 = 0 + 0.999 999 987 650 461 171 712;
  • 37) 0.999 999 987 650 461 171 712 × 2 = 1 + 0.999 999 975 300 922 343 424;
  • 38) 0.999 999 975 300 922 343 424 × 2 = 1 + 0.999 999 950 601 844 686 848;
  • 39) 0.999 999 950 601 844 686 848 × 2 = 1 + 0.999 999 901 203 689 373 696;
  • 40) 0.999 999 901 203 689 373 696 × 2 = 1 + 0.999 999 802 407 378 747 392;
  • 41) 0.999 999 802 407 378 747 392 × 2 = 1 + 0.999 999 604 814 757 494 784;
  • 42) 0.999 999 604 814 757 494 784 × 2 = 1 + 0.999 999 209 629 514 989 568;
  • 43) 0.999 999 209 629 514 989 568 × 2 = 1 + 0.999 998 419 259 029 979 136;
  • 44) 0.999 998 419 259 029 979 136 × 2 = 1 + 0.999 996 838 518 059 958 272;
  • 45) 0.999 996 838 518 059 958 272 × 2 = 1 + 0.999 993 677 036 119 916 544;
  • 46) 0.999 993 677 036 119 916 544 × 2 = 1 + 0.999 987 354 072 239 833 088;
  • 47) 0.999 987 354 072 239 833 088 × 2 = 1 + 0.999 974 708 144 479 666 176;
  • 48) 0.999 974 708 144 479 666 176 × 2 = 1 + 0.999 949 416 288 959 332 352;
  • 49) 0.999 949 416 288 959 332 352 × 2 = 1 + 0.999 898 832 577 918 664 704;
  • 50) 0.999 898 832 577 918 664 704 × 2 = 1 + 0.999 797 665 155 837 329 408;
  • 51) 0.999 797 665 155 837 329 408 × 2 = 1 + 0.999 595 330 311 674 658 816;
  • 52) 0.999 595 330 311 674 658 816 × 2 = 1 + 0.999 190 660 623 349 317 632;
  • 53) 0.999 190 660 623 349 317 632 × 2 = 1 + 0.998 381 321 246 698 635 264;
  • 54) 0.998 381 321 246 698 635 264 × 2 = 1 + 0.996 762 642 493 397 270 528;

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:


0.000 000 000 742 147 676 467(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 467(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:


0.000 000 000 742 147 676 467(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 -0.000 000 000 742 147 676 467 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