-0.000 000 000 742 147 676 652 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 652 1(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
-0.000 000 000 742 147 676 652 1(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:

|-0.000 000 000 742 147 676 652 1| = 0.000 000 000 742 147 676 652 1


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: 0.000 000 000 742 147 676 652 1.

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) 0.000 000 000 742 147 676 652 1 × 2 = 0 + 0.000 000 001 484 295 353 304 2;
  • 2) 0.000 000 001 484 295 353 304 2 × 2 = 0 + 0.000 000 002 968 590 706 608 4;
  • 3) 0.000 000 002 968 590 706 608 4 × 2 = 0 + 0.000 000 005 937 181 413 216 8;
  • 4) 0.000 000 005 937 181 413 216 8 × 2 = 0 + 0.000 000 011 874 362 826 433 6;
  • 5) 0.000 000 011 874 362 826 433 6 × 2 = 0 + 0.000 000 023 748 725 652 867 2;
  • 6) 0.000 000 023 748 725 652 867 2 × 2 = 0 + 0.000 000 047 497 451 305 734 4;
  • 7) 0.000 000 047 497 451 305 734 4 × 2 = 0 + 0.000 000 094 994 902 611 468 8;
  • 8) 0.000 000 094 994 902 611 468 8 × 2 = 0 + 0.000 000 189 989 805 222 937 6;
  • 9) 0.000 000 189 989 805 222 937 6 × 2 = 0 + 0.000 000 379 979 610 445 875 2;
  • 10) 0.000 000 379 979 610 445 875 2 × 2 = 0 + 0.000 000 759 959 220 891 750 4;
  • 11) 0.000 000 759 959 220 891 750 4 × 2 = 0 + 0.000 001 519 918 441 783 500 8;
  • 12) 0.000 001 519 918 441 783 500 8 × 2 = 0 + 0.000 003 039 836 883 567 001 6;
  • 13) 0.000 003 039 836 883 567 001 6 × 2 = 0 + 0.000 006 079 673 767 134 003 2;
  • 14) 0.000 006 079 673 767 134 003 2 × 2 = 0 + 0.000 012 159 347 534 268 006 4;
  • 15) 0.000 012 159 347 534 268 006 4 × 2 = 0 + 0.000 024 318 695 068 536 012 8;
  • 16) 0.000 024 318 695 068 536 012 8 × 2 = 0 + 0.000 048 637 390 137 072 025 6;
  • 17) 0.000 048 637 390 137 072 025 6 × 2 = 0 + 0.000 097 274 780 274 144 051 2;
  • 18) 0.000 097 274 780 274 144 051 2 × 2 = 0 + 0.000 194 549 560 548 288 102 4;
  • 19) 0.000 194 549 560 548 288 102 4 × 2 = 0 + 0.000 389 099 121 096 576 204 8;
  • 20) 0.000 389 099 121 096 576 204 8 × 2 = 0 + 0.000 778 198 242 193 152 409 6;
  • 21) 0.000 778 198 242 193 152 409 6 × 2 = 0 + 0.001 556 396 484 386 304 819 2;
  • 22) 0.001 556 396 484 386 304 819 2 × 2 = 0 + 0.003 112 792 968 772 609 638 4;
  • 23) 0.003 112 792 968 772 609 638 4 × 2 = 0 + 0.006 225 585 937 545 219 276 8;
  • 24) 0.006 225 585 937 545 219 276 8 × 2 = 0 + 0.012 451 171 875 090 438 553 6;
  • 25) 0.012 451 171 875 090 438 553 6 × 2 = 0 + 0.024 902 343 750 180 877 107 2;
  • 26) 0.024 902 343 750 180 877 107 2 × 2 = 0 + 0.049 804 687 500 361 754 214 4;
  • 27) 0.049 804 687 500 361 754 214 4 × 2 = 0 + 0.099 609 375 000 723 508 428 8;
  • 28) 0.099 609 375 000 723 508 428 8 × 2 = 0 + 0.199 218 750 001 447 016 857 6;
  • 29) 0.199 218 750 001 447 016 857 6 × 2 = 0 + 0.398 437 500 002 894 033 715 2;
  • 30) 0.398 437 500 002 894 033 715 2 × 2 = 0 + 0.796 875 000 005 788 067 430 4;
  • 31) 0.796 875 000 005 788 067 430 4 × 2 = 1 + 0.593 750 000 011 576 134 860 8;
  • 32) 0.593 750 000 011 576 134 860 8 × 2 = 1 + 0.187 500 000 023 152 269 721 6;
  • 33) 0.187 500 000 023 152 269 721 6 × 2 = 0 + 0.375 000 000 046 304 539 443 2;
  • 34) 0.375 000 000 046 304 539 443 2 × 2 = 0 + 0.750 000 000 092 609 078 886 4;
  • 35) 0.750 000 000 092 609 078 886 4 × 2 = 1 + 0.500 000 000 185 218 157 772 8;
  • 36) 0.500 000 000 185 218 157 772 8 × 2 = 1 + 0.000 000 000 370 436 315 545 6;
  • 37) 0.000 000 000 370 436 315 545 6 × 2 = 0 + 0.000 000 000 740 872 631 091 2;
  • 38) 0.000 000 000 740 872 631 091 2 × 2 = 0 + 0.000 000 001 481 745 262 182 4;
  • 39) 0.000 000 001 481 745 262 182 4 × 2 = 0 + 0.000 000 002 963 490 524 364 8;
  • 40) 0.000 000 002 963 490 524 364 8 × 2 = 0 + 0.000 000 005 926 981 048 729 6;
  • 41) 0.000 000 005 926 981 048 729 6 × 2 = 0 + 0.000 000 011 853 962 097 459 2;
  • 42) 0.000 000 011 853 962 097 459 2 × 2 = 0 + 0.000 000 023 707 924 194 918 4;
  • 43) 0.000 000 023 707 924 194 918 4 × 2 = 0 + 0.000 000 047 415 848 389 836 8;
  • 44) 0.000 000 047 415 848 389 836 8 × 2 = 0 + 0.000 000 094 831 696 779 673 6;
  • 45) 0.000 000 094 831 696 779 673 6 × 2 = 0 + 0.000 000 189 663 393 559 347 2;
  • 46) 0.000 000 189 663 393 559 347 2 × 2 = 0 + 0.000 000 379 326 787 118 694 4;
  • 47) 0.000 000 379 326 787 118 694 4 × 2 = 0 + 0.000 000 758 653 574 237 388 8;
  • 48) 0.000 000 758 653 574 237 388 8 × 2 = 0 + 0.000 001 517 307 148 474 777 6;
  • 49) 0.000 001 517 307 148 474 777 6 × 2 = 0 + 0.000 003 034 614 296 949 555 2;
  • 50) 0.000 003 034 614 296 949 555 2 × 2 = 0 + 0.000 006 069 228 593 899 110 4;
  • 51) 0.000 006 069 228 593 899 110 4 × 2 = 0 + 0.000 012 138 457 187 798 220 8;
  • 52) 0.000 012 138 457 187 798 220 8 × 2 = 0 + 0.000 024 276 914 375 596 441 6;
  • 53) 0.000 024 276 914 375 596 441 6 × 2 = 0 + 0.000 048 553 828 751 192 883 2;
  • 54) 0.000 048 553 828 751 192 883 2 × 2 = 0 + 0.000 097 107 657 502 385 766 4;

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:


0.000 000 000 742 147 676 652 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 652 1(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:


0.000 000 000 742 147 676 652 1(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 -0.000 000 000 742 147 676 652 1 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