-0.000 000 000 742 147 676 433 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 433(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
-0.000 000 000 742 147 676 433(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:

|-0.000 000 000 742 147 676 433| = 0.000 000 000 742 147 676 433


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: 0.000 000 000 742 147 676 433.

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) 0.000 000 000 742 147 676 433 × 2 = 0 + 0.000 000 001 484 295 352 866;
  • 2) 0.000 000 001 484 295 352 866 × 2 = 0 + 0.000 000 002 968 590 705 732;
  • 3) 0.000 000 002 968 590 705 732 × 2 = 0 + 0.000 000 005 937 181 411 464;
  • 4) 0.000 000 005 937 181 411 464 × 2 = 0 + 0.000 000 011 874 362 822 928;
  • 5) 0.000 000 011 874 362 822 928 × 2 = 0 + 0.000 000 023 748 725 645 856;
  • 6) 0.000 000 023 748 725 645 856 × 2 = 0 + 0.000 000 047 497 451 291 712;
  • 7) 0.000 000 047 497 451 291 712 × 2 = 0 + 0.000 000 094 994 902 583 424;
  • 8) 0.000 000 094 994 902 583 424 × 2 = 0 + 0.000 000 189 989 805 166 848;
  • 9) 0.000 000 189 989 805 166 848 × 2 = 0 + 0.000 000 379 979 610 333 696;
  • 10) 0.000 000 379 979 610 333 696 × 2 = 0 + 0.000 000 759 959 220 667 392;
  • 11) 0.000 000 759 959 220 667 392 × 2 = 0 + 0.000 001 519 918 441 334 784;
  • 12) 0.000 001 519 918 441 334 784 × 2 = 0 + 0.000 003 039 836 882 669 568;
  • 13) 0.000 003 039 836 882 669 568 × 2 = 0 + 0.000 006 079 673 765 339 136;
  • 14) 0.000 006 079 673 765 339 136 × 2 = 0 + 0.000 012 159 347 530 678 272;
  • 15) 0.000 012 159 347 530 678 272 × 2 = 0 + 0.000 024 318 695 061 356 544;
  • 16) 0.000 024 318 695 061 356 544 × 2 = 0 + 0.000 048 637 390 122 713 088;
  • 17) 0.000 048 637 390 122 713 088 × 2 = 0 + 0.000 097 274 780 245 426 176;
  • 18) 0.000 097 274 780 245 426 176 × 2 = 0 + 0.000 194 549 560 490 852 352;
  • 19) 0.000 194 549 560 490 852 352 × 2 = 0 + 0.000 389 099 120 981 704 704;
  • 20) 0.000 389 099 120 981 704 704 × 2 = 0 + 0.000 778 198 241 963 409 408;
  • 21) 0.000 778 198 241 963 409 408 × 2 = 0 + 0.001 556 396 483 926 818 816;
  • 22) 0.001 556 396 483 926 818 816 × 2 = 0 + 0.003 112 792 967 853 637 632;
  • 23) 0.003 112 792 967 853 637 632 × 2 = 0 + 0.006 225 585 935 707 275 264;
  • 24) 0.006 225 585 935 707 275 264 × 2 = 0 + 0.012 451 171 871 414 550 528;
  • 25) 0.012 451 171 871 414 550 528 × 2 = 0 + 0.024 902 343 742 829 101 056;
  • 26) 0.024 902 343 742 829 101 056 × 2 = 0 + 0.049 804 687 485 658 202 112;
  • 27) 0.049 804 687 485 658 202 112 × 2 = 0 + 0.099 609 374 971 316 404 224;
  • 28) 0.099 609 374 971 316 404 224 × 2 = 0 + 0.199 218 749 942 632 808 448;
  • 29) 0.199 218 749 942 632 808 448 × 2 = 0 + 0.398 437 499 885 265 616 896;
  • 30) 0.398 437 499 885 265 616 896 × 2 = 0 + 0.796 874 999 770 531 233 792;
  • 31) 0.796 874 999 770 531 233 792 × 2 = 1 + 0.593 749 999 541 062 467 584;
  • 32) 0.593 749 999 541 062 467 584 × 2 = 1 + 0.187 499 999 082 124 935 168;
  • 33) 0.187 499 999 082 124 935 168 × 2 = 0 + 0.374 999 998 164 249 870 336;
  • 34) 0.374 999 998 164 249 870 336 × 2 = 0 + 0.749 999 996 328 499 740 672;
  • 35) 0.749 999 996 328 499 740 672 × 2 = 1 + 0.499 999 992 656 999 481 344;
  • 36) 0.499 999 992 656 999 481 344 × 2 = 0 + 0.999 999 985 313 998 962 688;
  • 37) 0.999 999 985 313 998 962 688 × 2 = 1 + 0.999 999 970 627 997 925 376;
  • 38) 0.999 999 970 627 997 925 376 × 2 = 1 + 0.999 999 941 255 995 850 752;
  • 39) 0.999 999 941 255 995 850 752 × 2 = 1 + 0.999 999 882 511 991 701 504;
  • 40) 0.999 999 882 511 991 701 504 × 2 = 1 + 0.999 999 765 023 983 403 008;
  • 41) 0.999 999 765 023 983 403 008 × 2 = 1 + 0.999 999 530 047 966 806 016;
  • 42) 0.999 999 530 047 966 806 016 × 2 = 1 + 0.999 999 060 095 933 612 032;
  • 43) 0.999 999 060 095 933 612 032 × 2 = 1 + 0.999 998 120 191 867 224 064;
  • 44) 0.999 998 120 191 867 224 064 × 2 = 1 + 0.999 996 240 383 734 448 128;
  • 45) 0.999 996 240 383 734 448 128 × 2 = 1 + 0.999 992 480 767 468 896 256;
  • 46) 0.999 992 480 767 468 896 256 × 2 = 1 + 0.999 984 961 534 937 792 512;
  • 47) 0.999 984 961 534 937 792 512 × 2 = 1 + 0.999 969 923 069 875 585 024;
  • 48) 0.999 969 923 069 875 585 024 × 2 = 1 + 0.999 939 846 139 751 170 048;
  • 49) 0.999 939 846 139 751 170 048 × 2 = 1 + 0.999 879 692 279 502 340 096;
  • 50) 0.999 879 692 279 502 340 096 × 2 = 1 + 0.999 759 384 559 004 680 192;
  • 51) 0.999 759 384 559 004 680 192 × 2 = 1 + 0.999 518 769 118 009 360 384;
  • 52) 0.999 518 769 118 009 360 384 × 2 = 1 + 0.999 037 538 236 018 720 768;
  • 53) 0.999 037 538 236 018 720 768 × 2 = 1 + 0.998 075 076 472 037 441 536;
  • 54) 0.998 075 076 472 037 441 536 × 2 = 1 + 0.996 150 152 944 074 883 072;

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:


0.000 000 000 742 147 676 433(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 433(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:


0.000 000 000 742 147 676 433(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 -0.000 000 000 742 147 676 433 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