-0.000 000 000 742 147 676 621 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 621 7(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
-0.000 000 000 742 147 676 621 7(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:

|-0.000 000 000 742 147 676 621 7| = 0.000 000 000 742 147 676 621 7


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: 0.000 000 000 742 147 676 621 7.

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) 0.000 000 000 742 147 676 621 7 × 2 = 0 + 0.000 000 001 484 295 353 243 4;
  • 2) 0.000 000 001 484 295 353 243 4 × 2 = 0 + 0.000 000 002 968 590 706 486 8;
  • 3) 0.000 000 002 968 590 706 486 8 × 2 = 0 + 0.000 000 005 937 181 412 973 6;
  • 4) 0.000 000 005 937 181 412 973 6 × 2 = 0 + 0.000 000 011 874 362 825 947 2;
  • 5) 0.000 000 011 874 362 825 947 2 × 2 = 0 + 0.000 000 023 748 725 651 894 4;
  • 6) 0.000 000 023 748 725 651 894 4 × 2 = 0 + 0.000 000 047 497 451 303 788 8;
  • 7) 0.000 000 047 497 451 303 788 8 × 2 = 0 + 0.000 000 094 994 902 607 577 6;
  • 8) 0.000 000 094 994 902 607 577 6 × 2 = 0 + 0.000 000 189 989 805 215 155 2;
  • 9) 0.000 000 189 989 805 215 155 2 × 2 = 0 + 0.000 000 379 979 610 430 310 4;
  • 10) 0.000 000 379 979 610 430 310 4 × 2 = 0 + 0.000 000 759 959 220 860 620 8;
  • 11) 0.000 000 759 959 220 860 620 8 × 2 = 0 + 0.000 001 519 918 441 721 241 6;
  • 12) 0.000 001 519 918 441 721 241 6 × 2 = 0 + 0.000 003 039 836 883 442 483 2;
  • 13) 0.000 003 039 836 883 442 483 2 × 2 = 0 + 0.000 006 079 673 766 884 966 4;
  • 14) 0.000 006 079 673 766 884 966 4 × 2 = 0 + 0.000 012 159 347 533 769 932 8;
  • 15) 0.000 012 159 347 533 769 932 8 × 2 = 0 + 0.000 024 318 695 067 539 865 6;
  • 16) 0.000 024 318 695 067 539 865 6 × 2 = 0 + 0.000 048 637 390 135 079 731 2;
  • 17) 0.000 048 637 390 135 079 731 2 × 2 = 0 + 0.000 097 274 780 270 159 462 4;
  • 18) 0.000 097 274 780 270 159 462 4 × 2 = 0 + 0.000 194 549 560 540 318 924 8;
  • 19) 0.000 194 549 560 540 318 924 8 × 2 = 0 + 0.000 389 099 121 080 637 849 6;
  • 20) 0.000 389 099 121 080 637 849 6 × 2 = 0 + 0.000 778 198 242 161 275 699 2;
  • 21) 0.000 778 198 242 161 275 699 2 × 2 = 0 + 0.001 556 396 484 322 551 398 4;
  • 22) 0.001 556 396 484 322 551 398 4 × 2 = 0 + 0.003 112 792 968 645 102 796 8;
  • 23) 0.003 112 792 968 645 102 796 8 × 2 = 0 + 0.006 225 585 937 290 205 593 6;
  • 24) 0.006 225 585 937 290 205 593 6 × 2 = 0 + 0.012 451 171 874 580 411 187 2;
  • 25) 0.012 451 171 874 580 411 187 2 × 2 = 0 + 0.024 902 343 749 160 822 374 4;
  • 26) 0.024 902 343 749 160 822 374 4 × 2 = 0 + 0.049 804 687 498 321 644 748 8;
  • 27) 0.049 804 687 498 321 644 748 8 × 2 = 0 + 0.099 609 374 996 643 289 497 6;
  • 28) 0.099 609 374 996 643 289 497 6 × 2 = 0 + 0.199 218 749 993 286 578 995 2;
  • 29) 0.199 218 749 993 286 578 995 2 × 2 = 0 + 0.398 437 499 986 573 157 990 4;
  • 30) 0.398 437 499 986 573 157 990 4 × 2 = 0 + 0.796 874 999 973 146 315 980 8;
  • 31) 0.796 874 999 973 146 315 980 8 × 2 = 1 + 0.593 749 999 946 292 631 961 6;
  • 32) 0.593 749 999 946 292 631 961 6 × 2 = 1 + 0.187 499 999 892 585 263 923 2;
  • 33) 0.187 499 999 892 585 263 923 2 × 2 = 0 + 0.374 999 999 785 170 527 846 4;
  • 34) 0.374 999 999 785 170 527 846 4 × 2 = 0 + 0.749 999 999 570 341 055 692 8;
  • 35) 0.749 999 999 570 341 055 692 8 × 2 = 1 + 0.499 999 999 140 682 111 385 6;
  • 36) 0.499 999 999 140 682 111 385 6 × 2 = 0 + 0.999 999 998 281 364 222 771 2;
  • 37) 0.999 999 998 281 364 222 771 2 × 2 = 1 + 0.999 999 996 562 728 445 542 4;
  • 38) 0.999 999 996 562 728 445 542 4 × 2 = 1 + 0.999 999 993 125 456 891 084 8;
  • 39) 0.999 999 993 125 456 891 084 8 × 2 = 1 + 0.999 999 986 250 913 782 169 6;
  • 40) 0.999 999 986 250 913 782 169 6 × 2 = 1 + 0.999 999 972 501 827 564 339 2;
  • 41) 0.999 999 972 501 827 564 339 2 × 2 = 1 + 0.999 999 945 003 655 128 678 4;
  • 42) 0.999 999 945 003 655 128 678 4 × 2 = 1 + 0.999 999 890 007 310 257 356 8;
  • 43) 0.999 999 890 007 310 257 356 8 × 2 = 1 + 0.999 999 780 014 620 514 713 6;
  • 44) 0.999 999 780 014 620 514 713 6 × 2 = 1 + 0.999 999 560 029 241 029 427 2;
  • 45) 0.999 999 560 029 241 029 427 2 × 2 = 1 + 0.999 999 120 058 482 058 854 4;
  • 46) 0.999 999 120 058 482 058 854 4 × 2 = 1 + 0.999 998 240 116 964 117 708 8;
  • 47) 0.999 998 240 116 964 117 708 8 × 2 = 1 + 0.999 996 480 233 928 235 417 6;
  • 48) 0.999 996 480 233 928 235 417 6 × 2 = 1 + 0.999 992 960 467 856 470 835 2;
  • 49) 0.999 992 960 467 856 470 835 2 × 2 = 1 + 0.999 985 920 935 712 941 670 4;
  • 50) 0.999 985 920 935 712 941 670 4 × 2 = 1 + 0.999 971 841 871 425 883 340 8;
  • 51) 0.999 971 841 871 425 883 340 8 × 2 = 1 + 0.999 943 683 742 851 766 681 6;
  • 52) 0.999 943 683 742 851 766 681 6 × 2 = 1 + 0.999 887 367 485 703 533 363 2;
  • 53) 0.999 887 367 485 703 533 363 2 × 2 = 1 + 0.999 774 734 971 407 066 726 4;
  • 54) 0.999 774 734 971 407 066 726 4 × 2 = 1 + 0.999 549 469 942 814 133 452 8;

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:


0.000 000 000 742 147 676 621 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 621 7(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:


0.000 000 000 742 147 676 621 7(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 -0.000 000 000 742 147 676 621 7 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