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

Convert decimal -0.000 000 000 742 147 676 646 32(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 32(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 32| = 0.000 000 000 742 147 676 646 32


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 32.

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 32 × 2 = 0 + 0.000 000 001 484 295 353 292 64;
  • 2) 0.000 000 001 484 295 353 292 64 × 2 = 0 + 0.000 000 002 968 590 706 585 28;
  • 3) 0.000 000 002 968 590 706 585 28 × 2 = 0 + 0.000 000 005 937 181 413 170 56;
  • 4) 0.000 000 005 937 181 413 170 56 × 2 = 0 + 0.000 000 011 874 362 826 341 12;
  • 5) 0.000 000 011 874 362 826 341 12 × 2 = 0 + 0.000 000 023 748 725 652 682 24;
  • 6) 0.000 000 023 748 725 652 682 24 × 2 = 0 + 0.000 000 047 497 451 305 364 48;
  • 7) 0.000 000 047 497 451 305 364 48 × 2 = 0 + 0.000 000 094 994 902 610 728 96;
  • 8) 0.000 000 094 994 902 610 728 96 × 2 = 0 + 0.000 000 189 989 805 221 457 92;
  • 9) 0.000 000 189 989 805 221 457 92 × 2 = 0 + 0.000 000 379 979 610 442 915 84;
  • 10) 0.000 000 379 979 610 442 915 84 × 2 = 0 + 0.000 000 759 959 220 885 831 68;
  • 11) 0.000 000 759 959 220 885 831 68 × 2 = 0 + 0.000 001 519 918 441 771 663 36;
  • 12) 0.000 001 519 918 441 771 663 36 × 2 = 0 + 0.000 003 039 836 883 543 326 72;
  • 13) 0.000 003 039 836 883 543 326 72 × 2 = 0 + 0.000 006 079 673 767 086 653 44;
  • 14) 0.000 006 079 673 767 086 653 44 × 2 = 0 + 0.000 012 159 347 534 173 306 88;
  • 15) 0.000 012 159 347 534 173 306 88 × 2 = 0 + 0.000 024 318 695 068 346 613 76;
  • 16) 0.000 024 318 695 068 346 613 76 × 2 = 0 + 0.000 048 637 390 136 693 227 52;
  • 17) 0.000 048 637 390 136 693 227 52 × 2 = 0 + 0.000 097 274 780 273 386 455 04;
  • 18) 0.000 097 274 780 273 386 455 04 × 2 = 0 + 0.000 194 549 560 546 772 910 08;
  • 19) 0.000 194 549 560 546 772 910 08 × 2 = 0 + 0.000 389 099 121 093 545 820 16;
  • 20) 0.000 389 099 121 093 545 820 16 × 2 = 0 + 0.000 778 198 242 187 091 640 32;
  • 21) 0.000 778 198 242 187 091 640 32 × 2 = 0 + 0.001 556 396 484 374 183 280 64;
  • 22) 0.001 556 396 484 374 183 280 64 × 2 = 0 + 0.003 112 792 968 748 366 561 28;
  • 23) 0.003 112 792 968 748 366 561 28 × 2 = 0 + 0.006 225 585 937 496 733 122 56;
  • 24) 0.006 225 585 937 496 733 122 56 × 2 = 0 + 0.012 451 171 874 993 466 245 12;
  • 25) 0.012 451 171 874 993 466 245 12 × 2 = 0 + 0.024 902 343 749 986 932 490 24;
  • 26) 0.024 902 343 749 986 932 490 24 × 2 = 0 + 0.049 804 687 499 973 864 980 48;
  • 27) 0.049 804 687 499 973 864 980 48 × 2 = 0 + 0.099 609 374 999 947 729 960 96;
  • 28) 0.099 609 374 999 947 729 960 96 × 2 = 0 + 0.199 218 749 999 895 459 921 92;
  • 29) 0.199 218 749 999 895 459 921 92 × 2 = 0 + 0.398 437 499 999 790 919 843 84;
  • 30) 0.398 437 499 999 790 919 843 84 × 2 = 0 + 0.796 874 999 999 581 839 687 68;
  • 31) 0.796 874 999 999 581 839 687 68 × 2 = 1 + 0.593 749 999 999 163 679 375 36;
  • 32) 0.593 749 999 999 163 679 375 36 × 2 = 1 + 0.187 499 999 998 327 358 750 72;
  • 33) 0.187 499 999 998 327 358 750 72 × 2 = 0 + 0.374 999 999 996 654 717 501 44;
  • 34) 0.374 999 999 996 654 717 501 44 × 2 = 0 + 0.749 999 999 993 309 435 002 88;
  • 35) 0.749 999 999 993 309 435 002 88 × 2 = 1 + 0.499 999 999 986 618 870 005 76;
  • 36) 0.499 999 999 986 618 870 005 76 × 2 = 0 + 0.999 999 999 973 237 740 011 52;
  • 37) 0.999 999 999 973 237 740 011 52 × 2 = 1 + 0.999 999 999 946 475 480 023 04;
  • 38) 0.999 999 999 946 475 480 023 04 × 2 = 1 + 0.999 999 999 892 950 960 046 08;
  • 39) 0.999 999 999 892 950 960 046 08 × 2 = 1 + 0.999 999 999 785 901 920 092 16;
  • 40) 0.999 999 999 785 901 920 092 16 × 2 = 1 + 0.999 999 999 571 803 840 184 32;
  • 41) 0.999 999 999 571 803 840 184 32 × 2 = 1 + 0.999 999 999 143 607 680 368 64;
  • 42) 0.999 999 999 143 607 680 368 64 × 2 = 1 + 0.999 999 998 287 215 360 737 28;
  • 43) 0.999 999 998 287 215 360 737 28 × 2 = 1 + 0.999 999 996 574 430 721 474 56;
  • 44) 0.999 999 996 574 430 721 474 56 × 2 = 1 + 0.999 999 993 148 861 442 949 12;
  • 45) 0.999 999 993 148 861 442 949 12 × 2 = 1 + 0.999 999 986 297 722 885 898 24;
  • 46) 0.999 999 986 297 722 885 898 24 × 2 = 1 + 0.999 999 972 595 445 771 796 48;
  • 47) 0.999 999 972 595 445 771 796 48 × 2 = 1 + 0.999 999 945 190 891 543 592 96;
  • 48) 0.999 999 945 190 891 543 592 96 × 2 = 1 + 0.999 999 890 381 783 087 185 92;
  • 49) 0.999 999 890 381 783 087 185 92 × 2 = 1 + 0.999 999 780 763 566 174 371 84;
  • 50) 0.999 999 780 763 566 174 371 84 × 2 = 1 + 0.999 999 561 527 132 348 743 68;
  • 51) 0.999 999 561 527 132 348 743 68 × 2 = 1 + 0.999 999 123 054 264 697 487 36;
  • 52) 0.999 999 123 054 264 697 487 36 × 2 = 1 + 0.999 998 246 108 529 394 974 72;
  • 53) 0.999 998 246 108 529 394 974 72 × 2 = 1 + 0.999 996 492 217 058 789 949 44;
  • 54) 0.999 996 492 217 058 789 949 44 × 2 = 1 + 0.999 992 984 434 117 579 898 88;

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 32(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 32(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:


0.000 000 000 742 147 676 646 32(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 -0.000 000 000 742 147 676 646 32 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