-0.000 000 000 742 147 676 664 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 664 5(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
-0.000 000 000 742 147 676 664 5(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:

|-0.000 000 000 742 147 676 664 5| = 0.000 000 000 742 147 676 664 5


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: 0.000 000 000 742 147 676 664 5.

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) 0.000 000 000 742 147 676 664 5 × 2 = 0 + 0.000 000 001 484 295 353 329;
  • 2) 0.000 000 001 484 295 353 329 × 2 = 0 + 0.000 000 002 968 590 706 658;
  • 3) 0.000 000 002 968 590 706 658 × 2 = 0 + 0.000 000 005 937 181 413 316;
  • 4) 0.000 000 005 937 181 413 316 × 2 = 0 + 0.000 000 011 874 362 826 632;
  • 5) 0.000 000 011 874 362 826 632 × 2 = 0 + 0.000 000 023 748 725 653 264;
  • 6) 0.000 000 023 748 725 653 264 × 2 = 0 + 0.000 000 047 497 451 306 528;
  • 7) 0.000 000 047 497 451 306 528 × 2 = 0 + 0.000 000 094 994 902 613 056;
  • 8) 0.000 000 094 994 902 613 056 × 2 = 0 + 0.000 000 189 989 805 226 112;
  • 9) 0.000 000 189 989 805 226 112 × 2 = 0 + 0.000 000 379 979 610 452 224;
  • 10) 0.000 000 379 979 610 452 224 × 2 = 0 + 0.000 000 759 959 220 904 448;
  • 11) 0.000 000 759 959 220 904 448 × 2 = 0 + 0.000 001 519 918 441 808 896;
  • 12) 0.000 001 519 918 441 808 896 × 2 = 0 + 0.000 003 039 836 883 617 792;
  • 13) 0.000 003 039 836 883 617 792 × 2 = 0 + 0.000 006 079 673 767 235 584;
  • 14) 0.000 006 079 673 767 235 584 × 2 = 0 + 0.000 012 159 347 534 471 168;
  • 15) 0.000 012 159 347 534 471 168 × 2 = 0 + 0.000 024 318 695 068 942 336;
  • 16) 0.000 024 318 695 068 942 336 × 2 = 0 + 0.000 048 637 390 137 884 672;
  • 17) 0.000 048 637 390 137 884 672 × 2 = 0 + 0.000 097 274 780 275 769 344;
  • 18) 0.000 097 274 780 275 769 344 × 2 = 0 + 0.000 194 549 560 551 538 688;
  • 19) 0.000 194 549 560 551 538 688 × 2 = 0 + 0.000 389 099 121 103 077 376;
  • 20) 0.000 389 099 121 103 077 376 × 2 = 0 + 0.000 778 198 242 206 154 752;
  • 21) 0.000 778 198 242 206 154 752 × 2 = 0 + 0.001 556 396 484 412 309 504;
  • 22) 0.001 556 396 484 412 309 504 × 2 = 0 + 0.003 112 792 968 824 619 008;
  • 23) 0.003 112 792 968 824 619 008 × 2 = 0 + 0.006 225 585 937 649 238 016;
  • 24) 0.006 225 585 937 649 238 016 × 2 = 0 + 0.012 451 171 875 298 476 032;
  • 25) 0.012 451 171 875 298 476 032 × 2 = 0 + 0.024 902 343 750 596 952 064;
  • 26) 0.024 902 343 750 596 952 064 × 2 = 0 + 0.049 804 687 501 193 904 128;
  • 27) 0.049 804 687 501 193 904 128 × 2 = 0 + 0.099 609 375 002 387 808 256;
  • 28) 0.099 609 375 002 387 808 256 × 2 = 0 + 0.199 218 750 004 775 616 512;
  • 29) 0.199 218 750 004 775 616 512 × 2 = 0 + 0.398 437 500 009 551 233 024;
  • 30) 0.398 437 500 009 551 233 024 × 2 = 0 + 0.796 875 000 019 102 466 048;
  • 31) 0.796 875 000 019 102 466 048 × 2 = 1 + 0.593 750 000 038 204 932 096;
  • 32) 0.593 750 000 038 204 932 096 × 2 = 1 + 0.187 500 000 076 409 864 192;
  • 33) 0.187 500 000 076 409 864 192 × 2 = 0 + 0.375 000 000 152 819 728 384;
  • 34) 0.375 000 000 152 819 728 384 × 2 = 0 + 0.750 000 000 305 639 456 768;
  • 35) 0.750 000 000 305 639 456 768 × 2 = 1 + 0.500 000 000 611 278 913 536;
  • 36) 0.500 000 000 611 278 913 536 × 2 = 1 + 0.000 000 001 222 557 827 072;
  • 37) 0.000 000 001 222 557 827 072 × 2 = 0 + 0.000 000 002 445 115 654 144;
  • 38) 0.000 000 002 445 115 654 144 × 2 = 0 + 0.000 000 004 890 231 308 288;
  • 39) 0.000 000 004 890 231 308 288 × 2 = 0 + 0.000 000 009 780 462 616 576;
  • 40) 0.000 000 009 780 462 616 576 × 2 = 0 + 0.000 000 019 560 925 233 152;
  • 41) 0.000 000 019 560 925 233 152 × 2 = 0 + 0.000 000 039 121 850 466 304;
  • 42) 0.000 000 039 121 850 466 304 × 2 = 0 + 0.000 000 078 243 700 932 608;
  • 43) 0.000 000 078 243 700 932 608 × 2 = 0 + 0.000 000 156 487 401 865 216;
  • 44) 0.000 000 156 487 401 865 216 × 2 = 0 + 0.000 000 312 974 803 730 432;
  • 45) 0.000 000 312 974 803 730 432 × 2 = 0 + 0.000 000 625 949 607 460 864;
  • 46) 0.000 000 625 949 607 460 864 × 2 = 0 + 0.000 001 251 899 214 921 728;
  • 47) 0.000 001 251 899 214 921 728 × 2 = 0 + 0.000 002 503 798 429 843 456;
  • 48) 0.000 002 503 798 429 843 456 × 2 = 0 + 0.000 005 007 596 859 686 912;
  • 49) 0.000 005 007 596 859 686 912 × 2 = 0 + 0.000 010 015 193 719 373 824;
  • 50) 0.000 010 015 193 719 373 824 × 2 = 0 + 0.000 020 030 387 438 747 648;
  • 51) 0.000 020 030 387 438 747 648 × 2 = 0 + 0.000 040 060 774 877 495 296;
  • 52) 0.000 040 060 774 877 495 296 × 2 = 0 + 0.000 080 121 549 754 990 592;
  • 53) 0.000 080 121 549 754 990 592 × 2 = 0 + 0.000 160 243 099 509 981 184;
  • 54) 0.000 160 243 099 509 981 184 × 2 = 0 + 0.000 320 486 199 019 962 368;

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:


0.000 000 000 742 147 676 664 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 664 5(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:


0.000 000 000 742 147 676 664 5(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 -0.000 000 000 742 147 676 664 5 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