-0.000 000 000 742 147 609 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 609(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
-0.000 000 000 742 147 609(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:

|-0.000 000 000 742 147 609| = 0.000 000 000 742 147 609


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: 0.000 000 000 742 147 609.

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) 0.000 000 000 742 147 609 × 2 = 0 + 0.000 000 001 484 295 218;
  • 2) 0.000 000 001 484 295 218 × 2 = 0 + 0.000 000 002 968 590 436;
  • 3) 0.000 000 002 968 590 436 × 2 = 0 + 0.000 000 005 937 180 872;
  • 4) 0.000 000 005 937 180 872 × 2 = 0 + 0.000 000 011 874 361 744;
  • 5) 0.000 000 011 874 361 744 × 2 = 0 + 0.000 000 023 748 723 488;
  • 6) 0.000 000 023 748 723 488 × 2 = 0 + 0.000 000 047 497 446 976;
  • 7) 0.000 000 047 497 446 976 × 2 = 0 + 0.000 000 094 994 893 952;
  • 8) 0.000 000 094 994 893 952 × 2 = 0 + 0.000 000 189 989 787 904;
  • 9) 0.000 000 189 989 787 904 × 2 = 0 + 0.000 000 379 979 575 808;
  • 10) 0.000 000 379 979 575 808 × 2 = 0 + 0.000 000 759 959 151 616;
  • 11) 0.000 000 759 959 151 616 × 2 = 0 + 0.000 001 519 918 303 232;
  • 12) 0.000 001 519 918 303 232 × 2 = 0 + 0.000 003 039 836 606 464;
  • 13) 0.000 003 039 836 606 464 × 2 = 0 + 0.000 006 079 673 212 928;
  • 14) 0.000 006 079 673 212 928 × 2 = 0 + 0.000 012 159 346 425 856;
  • 15) 0.000 012 159 346 425 856 × 2 = 0 + 0.000 024 318 692 851 712;
  • 16) 0.000 024 318 692 851 712 × 2 = 0 + 0.000 048 637 385 703 424;
  • 17) 0.000 048 637 385 703 424 × 2 = 0 + 0.000 097 274 771 406 848;
  • 18) 0.000 097 274 771 406 848 × 2 = 0 + 0.000 194 549 542 813 696;
  • 19) 0.000 194 549 542 813 696 × 2 = 0 + 0.000 389 099 085 627 392;
  • 20) 0.000 389 099 085 627 392 × 2 = 0 + 0.000 778 198 171 254 784;
  • 21) 0.000 778 198 171 254 784 × 2 = 0 + 0.001 556 396 342 509 568;
  • 22) 0.001 556 396 342 509 568 × 2 = 0 + 0.003 112 792 685 019 136;
  • 23) 0.003 112 792 685 019 136 × 2 = 0 + 0.006 225 585 370 038 272;
  • 24) 0.006 225 585 370 038 272 × 2 = 0 + 0.012 451 170 740 076 544;
  • 25) 0.012 451 170 740 076 544 × 2 = 0 + 0.024 902 341 480 153 088;
  • 26) 0.024 902 341 480 153 088 × 2 = 0 + 0.049 804 682 960 306 176;
  • 27) 0.049 804 682 960 306 176 × 2 = 0 + 0.099 609 365 920 612 352;
  • 28) 0.099 609 365 920 612 352 × 2 = 0 + 0.199 218 731 841 224 704;
  • 29) 0.199 218 731 841 224 704 × 2 = 0 + 0.398 437 463 682 449 408;
  • 30) 0.398 437 463 682 449 408 × 2 = 0 + 0.796 874 927 364 898 816;
  • 31) 0.796 874 927 364 898 816 × 2 = 1 + 0.593 749 854 729 797 632;
  • 32) 0.593 749 854 729 797 632 × 2 = 1 + 0.187 499 709 459 595 264;
  • 33) 0.187 499 709 459 595 264 × 2 = 0 + 0.374 999 418 919 190 528;
  • 34) 0.374 999 418 919 190 528 × 2 = 0 + 0.749 998 837 838 381 056;
  • 35) 0.749 998 837 838 381 056 × 2 = 1 + 0.499 997 675 676 762 112;
  • 36) 0.499 997 675 676 762 112 × 2 = 0 + 0.999 995 351 353 524 224;
  • 37) 0.999 995 351 353 524 224 × 2 = 1 + 0.999 990 702 707 048 448;
  • 38) 0.999 990 702 707 048 448 × 2 = 1 + 0.999 981 405 414 096 896;
  • 39) 0.999 981 405 414 096 896 × 2 = 1 + 0.999 962 810 828 193 792;
  • 40) 0.999 962 810 828 193 792 × 2 = 1 + 0.999 925 621 656 387 584;
  • 41) 0.999 925 621 656 387 584 × 2 = 1 + 0.999 851 243 312 775 168;
  • 42) 0.999 851 243 312 775 168 × 2 = 1 + 0.999 702 486 625 550 336;
  • 43) 0.999 702 486 625 550 336 × 2 = 1 + 0.999 404 973 251 100 672;
  • 44) 0.999 404 973 251 100 672 × 2 = 1 + 0.998 809 946 502 201 344;
  • 45) 0.998 809 946 502 201 344 × 2 = 1 + 0.997 619 893 004 402 688;
  • 46) 0.997 619 893 004 402 688 × 2 = 1 + 0.995 239 786 008 805 376;
  • 47) 0.995 239 786 008 805 376 × 2 = 1 + 0.990 479 572 017 610 752;
  • 48) 0.990 479 572 017 610 752 × 2 = 1 + 0.980 959 144 035 221 504;
  • 49) 0.980 959 144 035 221 504 × 2 = 1 + 0.961 918 288 070 443 008;
  • 50) 0.961 918 288 070 443 008 × 2 = 1 + 0.923 836 576 140 886 016;
  • 51) 0.923 836 576 140 886 016 × 2 = 1 + 0.847 673 152 281 772 032;
  • 52) 0.847 673 152 281 772 032 × 2 = 1 + 0.695 346 304 563 544 064;
  • 53) 0.695 346 304 563 544 064 × 2 = 1 + 0.390 692 609 127 088 128;
  • 54) 0.390 692 609 127 088 128 × 2 = 0 + 0.781 385 218 254 176 256;

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:


0.000 000 000 742 147 609(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 609(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(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:


0.000 000 000 742 147 609(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2) × 20 =

1.1001 0111 1111 1111 1111 110(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 110


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 1110 =

100 1011 1111 1111 1111 1110


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 1110


Decimal number -0.000 000 000 742 147 609 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1110

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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