-0.000 000 000 742 147 676 641 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 641 8(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
-0.000 000 000 742 147 676 641 8(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:

|-0.000 000 000 742 147 676 641 8| = 0.000 000 000 742 147 676 641 8


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: 0.000 000 000 742 147 676 641 8.

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) 0.000 000 000 742 147 676 641 8 × 2 = 0 + 0.000 000 001 484 295 353 283 6;
  • 2) 0.000 000 001 484 295 353 283 6 × 2 = 0 + 0.000 000 002 968 590 706 567 2;
  • 3) 0.000 000 002 968 590 706 567 2 × 2 = 0 + 0.000 000 005 937 181 413 134 4;
  • 4) 0.000 000 005 937 181 413 134 4 × 2 = 0 + 0.000 000 011 874 362 826 268 8;
  • 5) 0.000 000 011 874 362 826 268 8 × 2 = 0 + 0.000 000 023 748 725 652 537 6;
  • 6) 0.000 000 023 748 725 652 537 6 × 2 = 0 + 0.000 000 047 497 451 305 075 2;
  • 7) 0.000 000 047 497 451 305 075 2 × 2 = 0 + 0.000 000 094 994 902 610 150 4;
  • 8) 0.000 000 094 994 902 610 150 4 × 2 = 0 + 0.000 000 189 989 805 220 300 8;
  • 9) 0.000 000 189 989 805 220 300 8 × 2 = 0 + 0.000 000 379 979 610 440 601 6;
  • 10) 0.000 000 379 979 610 440 601 6 × 2 = 0 + 0.000 000 759 959 220 881 203 2;
  • 11) 0.000 000 759 959 220 881 203 2 × 2 = 0 + 0.000 001 519 918 441 762 406 4;
  • 12) 0.000 001 519 918 441 762 406 4 × 2 = 0 + 0.000 003 039 836 883 524 812 8;
  • 13) 0.000 003 039 836 883 524 812 8 × 2 = 0 + 0.000 006 079 673 767 049 625 6;
  • 14) 0.000 006 079 673 767 049 625 6 × 2 = 0 + 0.000 012 159 347 534 099 251 2;
  • 15) 0.000 012 159 347 534 099 251 2 × 2 = 0 + 0.000 024 318 695 068 198 502 4;
  • 16) 0.000 024 318 695 068 198 502 4 × 2 = 0 + 0.000 048 637 390 136 397 004 8;
  • 17) 0.000 048 637 390 136 397 004 8 × 2 = 0 + 0.000 097 274 780 272 794 009 6;
  • 18) 0.000 097 274 780 272 794 009 6 × 2 = 0 + 0.000 194 549 560 545 588 019 2;
  • 19) 0.000 194 549 560 545 588 019 2 × 2 = 0 + 0.000 389 099 121 091 176 038 4;
  • 20) 0.000 389 099 121 091 176 038 4 × 2 = 0 + 0.000 778 198 242 182 352 076 8;
  • 21) 0.000 778 198 242 182 352 076 8 × 2 = 0 + 0.001 556 396 484 364 704 153 6;
  • 22) 0.001 556 396 484 364 704 153 6 × 2 = 0 + 0.003 112 792 968 729 408 307 2;
  • 23) 0.003 112 792 968 729 408 307 2 × 2 = 0 + 0.006 225 585 937 458 816 614 4;
  • 24) 0.006 225 585 937 458 816 614 4 × 2 = 0 + 0.012 451 171 874 917 633 228 8;
  • 25) 0.012 451 171 874 917 633 228 8 × 2 = 0 + 0.024 902 343 749 835 266 457 6;
  • 26) 0.024 902 343 749 835 266 457 6 × 2 = 0 + 0.049 804 687 499 670 532 915 2;
  • 27) 0.049 804 687 499 670 532 915 2 × 2 = 0 + 0.099 609 374 999 341 065 830 4;
  • 28) 0.099 609 374 999 341 065 830 4 × 2 = 0 + 0.199 218 749 998 682 131 660 8;
  • 29) 0.199 218 749 998 682 131 660 8 × 2 = 0 + 0.398 437 499 997 364 263 321 6;
  • 30) 0.398 437 499 997 364 263 321 6 × 2 = 0 + 0.796 874 999 994 728 526 643 2;
  • 31) 0.796 874 999 994 728 526 643 2 × 2 = 1 + 0.593 749 999 989 457 053 286 4;
  • 32) 0.593 749 999 989 457 053 286 4 × 2 = 1 + 0.187 499 999 978 914 106 572 8;
  • 33) 0.187 499 999 978 914 106 572 8 × 2 = 0 + 0.374 999 999 957 828 213 145 6;
  • 34) 0.374 999 999 957 828 213 145 6 × 2 = 0 + 0.749 999 999 915 656 426 291 2;
  • 35) 0.749 999 999 915 656 426 291 2 × 2 = 1 + 0.499 999 999 831 312 852 582 4;
  • 36) 0.499 999 999 831 312 852 582 4 × 2 = 0 + 0.999 999 999 662 625 705 164 8;
  • 37) 0.999 999 999 662 625 705 164 8 × 2 = 1 + 0.999 999 999 325 251 410 329 6;
  • 38) 0.999 999 999 325 251 410 329 6 × 2 = 1 + 0.999 999 998 650 502 820 659 2;
  • 39) 0.999 999 998 650 502 820 659 2 × 2 = 1 + 0.999 999 997 301 005 641 318 4;
  • 40) 0.999 999 997 301 005 641 318 4 × 2 = 1 + 0.999 999 994 602 011 282 636 8;
  • 41) 0.999 999 994 602 011 282 636 8 × 2 = 1 + 0.999 999 989 204 022 565 273 6;
  • 42) 0.999 999 989 204 022 565 273 6 × 2 = 1 + 0.999 999 978 408 045 130 547 2;
  • 43) 0.999 999 978 408 045 130 547 2 × 2 = 1 + 0.999 999 956 816 090 261 094 4;
  • 44) 0.999 999 956 816 090 261 094 4 × 2 = 1 + 0.999 999 913 632 180 522 188 8;
  • 45) 0.999 999 913 632 180 522 188 8 × 2 = 1 + 0.999 999 827 264 361 044 377 6;
  • 46) 0.999 999 827 264 361 044 377 6 × 2 = 1 + 0.999 999 654 528 722 088 755 2;
  • 47) 0.999 999 654 528 722 088 755 2 × 2 = 1 + 0.999 999 309 057 444 177 510 4;
  • 48) 0.999 999 309 057 444 177 510 4 × 2 = 1 + 0.999 998 618 114 888 355 020 8;
  • 49) 0.999 998 618 114 888 355 020 8 × 2 = 1 + 0.999 997 236 229 776 710 041 6;
  • 50) 0.999 997 236 229 776 710 041 6 × 2 = 1 + 0.999 994 472 459 553 420 083 2;
  • 51) 0.999 994 472 459 553 420 083 2 × 2 = 1 + 0.999 988 944 919 106 840 166 4;
  • 52) 0.999 988 944 919 106 840 166 4 × 2 = 1 + 0.999 977 889 838 213 680 332 8;
  • 53) 0.999 977 889 838 213 680 332 8 × 2 = 1 + 0.999 955 779 676 427 360 665 6;
  • 54) 0.999 955 779 676 427 360 665 6 × 2 = 1 + 0.999 911 559 352 854 721 331 2;

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:


0.000 000 000 742 147 676 641 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 641 8(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:


0.000 000 000 742 147 676 641 8(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 -0.000 000 000 742 147 676 641 8 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