-0.000 000 000 742 147 676 646 849 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 849(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
-0.000 000 000 742 147 676 646 849(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:

|-0.000 000 000 742 147 676 646 849| = 0.000 000 000 742 147 676 646 849


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: 0.000 000 000 742 147 676 646 849.

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) 0.000 000 000 742 147 676 646 849 × 2 = 0 + 0.000 000 001 484 295 353 293 698;
  • 2) 0.000 000 001 484 295 353 293 698 × 2 = 0 + 0.000 000 002 968 590 706 587 396;
  • 3) 0.000 000 002 968 590 706 587 396 × 2 = 0 + 0.000 000 005 937 181 413 174 792;
  • 4) 0.000 000 005 937 181 413 174 792 × 2 = 0 + 0.000 000 011 874 362 826 349 584;
  • 5) 0.000 000 011 874 362 826 349 584 × 2 = 0 + 0.000 000 023 748 725 652 699 168;
  • 6) 0.000 000 023 748 725 652 699 168 × 2 = 0 + 0.000 000 047 497 451 305 398 336;
  • 7) 0.000 000 047 497 451 305 398 336 × 2 = 0 + 0.000 000 094 994 902 610 796 672;
  • 8) 0.000 000 094 994 902 610 796 672 × 2 = 0 + 0.000 000 189 989 805 221 593 344;
  • 9) 0.000 000 189 989 805 221 593 344 × 2 = 0 + 0.000 000 379 979 610 443 186 688;
  • 10) 0.000 000 379 979 610 443 186 688 × 2 = 0 + 0.000 000 759 959 220 886 373 376;
  • 11) 0.000 000 759 959 220 886 373 376 × 2 = 0 + 0.000 001 519 918 441 772 746 752;
  • 12) 0.000 001 519 918 441 772 746 752 × 2 = 0 + 0.000 003 039 836 883 545 493 504;
  • 13) 0.000 003 039 836 883 545 493 504 × 2 = 0 + 0.000 006 079 673 767 090 987 008;
  • 14) 0.000 006 079 673 767 090 987 008 × 2 = 0 + 0.000 012 159 347 534 181 974 016;
  • 15) 0.000 012 159 347 534 181 974 016 × 2 = 0 + 0.000 024 318 695 068 363 948 032;
  • 16) 0.000 024 318 695 068 363 948 032 × 2 = 0 + 0.000 048 637 390 136 727 896 064;
  • 17) 0.000 048 637 390 136 727 896 064 × 2 = 0 + 0.000 097 274 780 273 455 792 128;
  • 18) 0.000 097 274 780 273 455 792 128 × 2 = 0 + 0.000 194 549 560 546 911 584 256;
  • 19) 0.000 194 549 560 546 911 584 256 × 2 = 0 + 0.000 389 099 121 093 823 168 512;
  • 20) 0.000 389 099 121 093 823 168 512 × 2 = 0 + 0.000 778 198 242 187 646 337 024;
  • 21) 0.000 778 198 242 187 646 337 024 × 2 = 0 + 0.001 556 396 484 375 292 674 048;
  • 22) 0.001 556 396 484 375 292 674 048 × 2 = 0 + 0.003 112 792 968 750 585 348 096;
  • 23) 0.003 112 792 968 750 585 348 096 × 2 = 0 + 0.006 225 585 937 501 170 696 192;
  • 24) 0.006 225 585 937 501 170 696 192 × 2 = 0 + 0.012 451 171 875 002 341 392 384;
  • 25) 0.012 451 171 875 002 341 392 384 × 2 = 0 + 0.024 902 343 750 004 682 784 768;
  • 26) 0.024 902 343 750 004 682 784 768 × 2 = 0 + 0.049 804 687 500 009 365 569 536;
  • 27) 0.049 804 687 500 009 365 569 536 × 2 = 0 + 0.099 609 375 000 018 731 139 072;
  • 28) 0.099 609 375 000 018 731 139 072 × 2 = 0 + 0.199 218 750 000 037 462 278 144;
  • 29) 0.199 218 750 000 037 462 278 144 × 2 = 0 + 0.398 437 500 000 074 924 556 288;
  • 30) 0.398 437 500 000 074 924 556 288 × 2 = 0 + 0.796 875 000 000 149 849 112 576;
  • 31) 0.796 875 000 000 149 849 112 576 × 2 = 1 + 0.593 750 000 000 299 698 225 152;
  • 32) 0.593 750 000 000 299 698 225 152 × 2 = 1 + 0.187 500 000 000 599 396 450 304;
  • 33) 0.187 500 000 000 599 396 450 304 × 2 = 0 + 0.375 000 000 001 198 792 900 608;
  • 34) 0.375 000 000 001 198 792 900 608 × 2 = 0 + 0.750 000 000 002 397 585 801 216;
  • 35) 0.750 000 000 002 397 585 801 216 × 2 = 1 + 0.500 000 000 004 795 171 602 432;
  • 36) 0.500 000 000 004 795 171 602 432 × 2 = 1 + 0.000 000 000 009 590 343 204 864;
  • 37) 0.000 000 000 009 590 343 204 864 × 2 = 0 + 0.000 000 000 019 180 686 409 728;
  • 38) 0.000 000 000 019 180 686 409 728 × 2 = 0 + 0.000 000 000 038 361 372 819 456;
  • 39) 0.000 000 000 038 361 372 819 456 × 2 = 0 + 0.000 000 000 076 722 745 638 912;
  • 40) 0.000 000 000 076 722 745 638 912 × 2 = 0 + 0.000 000 000 153 445 491 277 824;
  • 41) 0.000 000 000 153 445 491 277 824 × 2 = 0 + 0.000 000 000 306 890 982 555 648;
  • 42) 0.000 000 000 306 890 982 555 648 × 2 = 0 + 0.000 000 000 613 781 965 111 296;
  • 43) 0.000 000 000 613 781 965 111 296 × 2 = 0 + 0.000 000 001 227 563 930 222 592;
  • 44) 0.000 000 001 227 563 930 222 592 × 2 = 0 + 0.000 000 002 455 127 860 445 184;
  • 45) 0.000 000 002 455 127 860 445 184 × 2 = 0 + 0.000 000 004 910 255 720 890 368;
  • 46) 0.000 000 004 910 255 720 890 368 × 2 = 0 + 0.000 000 009 820 511 441 780 736;
  • 47) 0.000 000 009 820 511 441 780 736 × 2 = 0 + 0.000 000 019 641 022 883 561 472;
  • 48) 0.000 000 019 641 022 883 561 472 × 2 = 0 + 0.000 000 039 282 045 767 122 944;
  • 49) 0.000 000 039 282 045 767 122 944 × 2 = 0 + 0.000 000 078 564 091 534 245 888;
  • 50) 0.000 000 078 564 091 534 245 888 × 2 = 0 + 0.000 000 157 128 183 068 491 776;
  • 51) 0.000 000 157 128 183 068 491 776 × 2 = 0 + 0.000 000 314 256 366 136 983 552;
  • 52) 0.000 000 314 256 366 136 983 552 × 2 = 0 + 0.000 000 628 512 732 273 967 104;
  • 53) 0.000 000 628 512 732 273 967 104 × 2 = 0 + 0.000 001 257 025 464 547 934 208;
  • 54) 0.000 001 257 025 464 547 934 208 × 2 = 0 + 0.000 002 514 050 929 095 868 416;

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:


0.000 000 000 742 147 676 646 849(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 849(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 849(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 849 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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