# Convert 0.000 000 000 000 000 000 016 04 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

## 0.000 000 000 000 000 000 016 04(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

### 1. First, convert to the binary (base 2) the integer part: 0. Divide the number repeatedly by 2.

#### We stop when we get a quotient that is equal to zero.

• division = quotient + remainder;
• 0 ÷ 2 = 0 + 0;

### 3. Convert to the binary (base 2) the fractional part: 0.000 000 000 000 000 000 016 04.

#### Stop when we get a fractional part that is equal to zero.

• #) multiplying = integer + fractional part;
• 1) 0.000 000 000 000 000 000 016 04 × 2 = 0 + 0.000 000 000 000 000 000 032 08;
• 2) 0.000 000 000 000 000 000 032 08 × 2 = 0 + 0.000 000 000 000 000 000 064 16;
• 3) 0.000 000 000 000 000 000 064 16 × 2 = 0 + 0.000 000 000 000 000 000 128 32;
• 4) 0.000 000 000 000 000 000 128 32 × 2 = 0 + 0.000 000 000 000 000 000 256 64;
• 5) 0.000 000 000 000 000 000 256 64 × 2 = 0 + 0.000 000 000 000 000 000 513 28;
• 6) 0.000 000 000 000 000 000 513 28 × 2 = 0 + 0.000 000 000 000 000 001 026 56;
• 7) 0.000 000 000 000 000 001 026 56 × 2 = 0 + 0.000 000 000 000 000 002 053 12;
• 8) 0.000 000 000 000 000 002 053 12 × 2 = 0 + 0.000 000 000 000 000 004 106 24;
• 9) 0.000 000 000 000 000 004 106 24 × 2 = 0 + 0.000 000 000 000 000 008 212 48;
• 10) 0.000 000 000 000 000 008 212 48 × 2 = 0 + 0.000 000 000 000 000 016 424 96;
• 11) 0.000 000 000 000 000 016 424 96 × 2 = 0 + 0.000 000 000 000 000 032 849 92;
• 12) 0.000 000 000 000 000 032 849 92 × 2 = 0 + 0.000 000 000 000 000 065 699 84;
• 13) 0.000 000 000 000 000 065 699 84 × 2 = 0 + 0.000 000 000 000 000 131 399 68;
• 14) 0.000 000 000 000 000 131 399 68 × 2 = 0 + 0.000 000 000 000 000 262 799 36;
• 15) 0.000 000 000 000 000 262 799 36 × 2 = 0 + 0.000 000 000 000 000 525 598 72;
• 16) 0.000 000 000 000 000 525 598 72 × 2 = 0 + 0.000 000 000 000 001 051 197 44;
• 17) 0.000 000 000 000 001 051 197 44 × 2 = 0 + 0.000 000 000 000 002 102 394 88;
• 18) 0.000 000 000 000 002 102 394 88 × 2 = 0 + 0.000 000 000 000 004 204 789 76;
• 19) 0.000 000 000 000 004 204 789 76 × 2 = 0 + 0.000 000 000 000 008 409 579 52;
• 20) 0.000 000 000 000 008 409 579 52 × 2 = 0 + 0.000 000 000 000 016 819 159 04;
• 21) 0.000 000 000 000 016 819 159 04 × 2 = 0 + 0.000 000 000 000 033 638 318 08;
• 22) 0.000 000 000 000 033 638 318 08 × 2 = 0 + 0.000 000 000 000 067 276 636 16;
• 23) 0.000 000 000 000 067 276 636 16 × 2 = 0 + 0.000 000 000 000 134 553 272 32;
• 24) 0.000 000 000 000 134 553 272 32 × 2 = 0 + 0.000 000 000 000 269 106 544 64;
• 25) 0.000 000 000 000 269 106 544 64 × 2 = 0 + 0.000 000 000 000 538 213 089 28;
• 26) 0.000 000 000 000 538 213 089 28 × 2 = 0 + 0.000 000 000 001 076 426 178 56;
• 27) 0.000 000 000 001 076 426 178 56 × 2 = 0 + 0.000 000 000 002 152 852 357 12;
• 28) 0.000 000 000 002 152 852 357 12 × 2 = 0 + 0.000 000 000 004 305 704 714 24;
• 29) 0.000 000 000 004 305 704 714 24 × 2 = 0 + 0.000 000 000 008 611 409 428 48;
• 30) 0.000 000 000 008 611 409 428 48 × 2 = 0 + 0.000 000 000 017 222 818 856 96;
• 31) 0.000 000 000 017 222 818 856 96 × 2 = 0 + 0.000 000 000 034 445 637 713 92;
• 32) 0.000 000 000 034 445 637 713 92 × 2 = 0 + 0.000 000 000 068 891 275 427 84;
• 33) 0.000 000 000 068 891 275 427 84 × 2 = 0 + 0.000 000 000 137 782 550 855 68;
• 34) 0.000 000 000 137 782 550 855 68 × 2 = 0 + 0.000 000 000 275 565 101 711 36;
• 35) 0.000 000 000 275 565 101 711 36 × 2 = 0 + 0.000 000 000 551 130 203 422 72;
• 36) 0.000 000 000 551 130 203 422 72 × 2 = 0 + 0.000 000 001 102 260 406 845 44;
• 37) 0.000 000 001 102 260 406 845 44 × 2 = 0 + 0.000 000 002 204 520 813 690 88;
• 38) 0.000 000 002 204 520 813 690 88 × 2 = 0 + 0.000 000 004 409 041 627 381 76;
• 39) 0.000 000 004 409 041 627 381 76 × 2 = 0 + 0.000 000 008 818 083 254 763 52;
• 40) 0.000 000 008 818 083 254 763 52 × 2 = 0 + 0.000 000 017 636 166 509 527 04;
• 41) 0.000 000 017 636 166 509 527 04 × 2 = 0 + 0.000 000 035 272 333 019 054 08;
• 42) 0.000 000 035 272 333 019 054 08 × 2 = 0 + 0.000 000 070 544 666 038 108 16;
• 43) 0.000 000 070 544 666 038 108 16 × 2 = 0 + 0.000 000 141 089 332 076 216 32;
• 44) 0.000 000 141 089 332 076 216 32 × 2 = 0 + 0.000 000 282 178 664 152 432 64;
• 45) 0.000 000 282 178 664 152 432 64 × 2 = 0 + 0.000 000 564 357 328 304 865 28;
• 46) 0.000 000 564 357 328 304 865 28 × 2 = 0 + 0.000 001 128 714 656 609 730 56;
• 47) 0.000 001 128 714 656 609 730 56 × 2 = 0 + 0.000 002 257 429 313 219 461 12;
• 48) 0.000 002 257 429 313 219 461 12 × 2 = 0 + 0.000 004 514 858 626 438 922 24;
• 49) 0.000 004 514 858 626 438 922 24 × 2 = 0 + 0.000 009 029 717 252 877 844 48;
• 50) 0.000 009 029 717 252 877 844 48 × 2 = 0 + 0.000 018 059 434 505 755 688 96;
• 51) 0.000 018 059 434 505 755 688 96 × 2 = 0 + 0.000 036 118 869 011 511 377 92;
• 52) 0.000 036 118 869 011 511 377 92 × 2 = 0 + 0.000 072 237 738 023 022 755 84;
• 53) 0.000 072 237 738 023 022 755 84 × 2 = 0 + 0.000 144 475 476 046 045 511 68;
• 54) 0.000 144 475 476 046 045 511 68 × 2 = 0 + 0.000 288 950 952 092 091 023 36;
• 55) 0.000 288 950 952 092 091 023 36 × 2 = 0 + 0.000 577 901 904 184 182 046 72;
• 56) 0.000 577 901 904 184 182 046 72 × 2 = 0 + 0.001 155 803 808 368 364 093 44;
• 57) 0.001 155 803 808 368 364 093 44 × 2 = 0 + 0.002 311 607 616 736 728 186 88;
• 58) 0.002 311 607 616 736 728 186 88 × 2 = 0 + 0.004 623 215 233 473 456 373 76;
• 59) 0.004 623 215 233 473 456 373 76 × 2 = 0 + 0.009 246 430 466 946 912 747 52;
• 60) 0.009 246 430 466 946 912 747 52 × 2 = 0 + 0.018 492 860 933 893 825 495 04;
• 61) 0.018 492 860 933 893 825 495 04 × 2 = 0 + 0.036 985 721 867 787 650 990 08;
• 62) 0.036 985 721 867 787 650 990 08 × 2 = 0 + 0.073 971 443 735 575 301 980 16;
• 63) 0.073 971 443 735 575 301 980 16 × 2 = 0 + 0.147 942 887 471 150 603 960 32;
• 64) 0.147 942 887 471 150 603 960 32 × 2 = 0 + 0.295 885 774 942 301 207 920 64;
• 65) 0.295 885 774 942 301 207 920 64 × 2 = 0 + 0.591 771 549 884 602 415 841 28;
• 66) 0.591 771 549 884 602 415 841 28 × 2 = 1 + 0.183 543 099 769 204 831 682 56;
• 67) 0.183 543 099 769 204 831 682 56 × 2 = 0 + 0.367 086 199 538 409 663 365 12;
• 68) 0.367 086 199 538 409 663 365 12 × 2 = 0 + 0.734 172 399 076 819 326 730 24;
• 69) 0.734 172 399 076 819 326 730 24 × 2 = 1 + 0.468 344 798 153 638 653 460 48;
• 70) 0.468 344 798 153 638 653 460 48 × 2 = 0 + 0.936 689 596 307 277 306 920 96;
• 71) 0.936 689 596 307 277 306 920 96 × 2 = 1 + 0.873 379 192 614 554 613 841 92;
• 72) 0.873 379 192 614 554 613 841 92 × 2 = 1 + 0.746 758 385 229 109 227 683 84;
• 73) 0.746 758 385 229 109 227 683 84 × 2 = 1 + 0.493 516 770 458 218 455 367 68;
• 74) 0.493 516 770 458 218 455 367 68 × 2 = 0 + 0.987 033 540 916 436 910 735 36;
• 75) 0.987 033 540 916 436 910 735 36 × 2 = 1 + 0.974 067 081 832 873 821 470 72;
• 76) 0.974 067 081 832 873 821 470 72 × 2 = 1 + 0.948 134 163 665 747 642 941 44;
• 77) 0.948 134 163 665 747 642 941 44 × 2 = 1 + 0.896 268 327 331 495 285 882 88;
• 78) 0.896 268 327 331 495 285 882 88 × 2 = 1 + 0.792 536 654 662 990 571 765 76;
• 79) 0.792 536 654 662 990 571 765 76 × 2 = 1 + 0.585 073 309 325 981 143 531 52;
• 80) 0.585 073 309 325 981 143 531 52 × 2 = 1 + 0.170 146 618 651 962 287 063 04;
• 81) 0.170 146 618 651 962 287 063 04 × 2 = 0 + 0.340 293 237 303 924 574 126 08;
• 82) 0.340 293 237 303 924 574 126 08 × 2 = 0 + 0.680 586 474 607 849 148 252 16;
• 83) 0.680 586 474 607 849 148 252 16 × 2 = 1 + 0.361 172 949 215 698 296 504 32;
• 84) 0.361 172 949 215 698 296 504 32 × 2 = 0 + 0.722 345 898 431 396 593 008 64;
• 85) 0.722 345 898 431 396 593 008 64 × 2 = 1 + 0.444 691 796 862 793 186 017 28;
• 86) 0.444 691 796 862 793 186 017 28 × 2 = 0 + 0.889 383 593 725 586 372 034 56;
• 87) 0.889 383 593 725 586 372 034 56 × 2 = 1 + 0.778 767 187 451 172 744 069 12;
• 88) 0.778 767 187 451 172 744 069 12 × 2 = 1 + 0.557 534 374 902 345 488 138 24;
• 89) 0.557 534 374 902 345 488 138 24 × 2 = 1 + 0.115 068 749 804 690 976 276 48;

### 9. 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;
• 61 ÷ 2 = 30 + 1;
• 30 ÷ 2 = 15 + 0;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

## Number 0.000 000 000 000 000 000 016 04 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point: 0 - 0011 1101 - 001 0111 0111 1110 0101 0111

(32 bits IEEE 754)

• 0

31

• 0

30
• 0

29
• 1

28
• 1

27
• 1

26
• 1

25
• 0

24
• 1

23

• 0

22
• 0

21
• 1

20
• 0

19
• 1

18
• 1

17
• 1

16
• 0

15
• 1

14
• 1

13
• 1

12
• 1

11
• 1

10
• 1

9
• 0

8
• 0

7
• 1

6
• 0

5
• 1

4
• 0

3
• 1

2
• 1

1
• 1

0

## Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

 0.000 000 000 000 000 000 016 04 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:09 UTC (GMT) 1 011 010 101 111 001 035 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:08 UTC (GMT) -510.1 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:08 UTC (GMT) 1 111 111 111 111 111 110 000 000 000 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:08 UTC (GMT) 117.15 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:08 UTC (GMT) 399 076 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:08 UTC (GMT) 0.029 25 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:07 UTC (GMT) -4 652 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:07 UTC (GMT) 121 123.28 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:07 UTC (GMT) 10 001 001 010 111 010 009 998 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:07 UTC (GMT) 0.560 5 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:07 UTC (GMT) 1 000 110 110 109 999 999 999 999 999 977 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:06 UTC (GMT) 1 011 110 000 999 999 999 999 999 999 993 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 09:06 UTC (GMT) All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

## 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:
• 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:

|-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)

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